Electrochemistry
A volume of x mL of 5 M NaHCO3 solution was mixed with 10 mL of 2 M H2CO3 solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x = ______ mL (nearest integer).
Sn(s) | Sn(OH)62− (0.5 M) | HSnO2− (0.05 M) | OH− | Bi2O3(s) | Bi(s)
Consider up to one place of decimal for intermediate calculations
$\left[\begin{array}{ll}\text { Given: } & E_{Sn\left( {OH} \right)_6^{2 - } |HSnO_2^ -}^o = - 0.9V \\ & \mathrm{E}^{\mathrm{o}}{ }_{\mathrm{Bi}_2 \mathrm{O}_3 \mid \mathrm{Bi}}=-0.44 \mathrm{~V} \\ & \mathrm{pKa}_{\left(\mathrm{H}_2 \mathrm{CO}_3\right)}=6.11 \\ & \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V} \\ & \text { Antilog }(1.29)=19.5\end{array}\right]$
Explanation:
For the half-cells (25 °C):
(1) Sn(IV)/Sn(II) couple (given $E^\circ=-0.90\text{ V}$)
Balanced reduction in basic medium:
$ \mathrm{Sn(OH)_6^{2-}+2e^- \rightarrow HSnO_2^-+3OH^-+H_2O} $
$ E_{\text{Sn}}=E^\circ_{\text{Sn}}-\frac{0.059}{2}\log\!\left(\frac{[HSnO_2^-][OH^-]^3}{[Sn(OH)_6^{2-}]}\right) $
Given $[Sn(OH)_6^{2-}]=0.5$, $[HSnO_2^-]=0.05$:
$ \frac{[HSnO_2^-]}{[Sn(OH)_6^{2-}]}=\frac{0.05}{0.5}=0.1 $
(2) Bi$_2$O$_3$/Bi couple (given $E^\circ=-0.44\text{ V}$)
Reduction:
$ \mathrm{Bi_2O_3(s)+3H_2O+6e^- \rightarrow 2Bi(s)+6OH^-} $
$ E_{\text{Bi}}=E^\circ_{\text{Bi}}-\frac{0.059}{6}\log([OH^-]^6)=E^\circ_{\text{Bi}}-0.059\log[OH^-] $
Bi has higher $E^\circ$, so it is the cathode. Thus
$ E_{\text{cell}}=E_{\text{cathode}}-E_{\text{anode}}=E_{\text{Bi}}-E_{\text{Sn}} $
Substitute and simplify:
$ E_{\text{cell}}=\left(-0.44-0.059\log[OH^-]\right)-\left(-0.90-\frac{0.059}{2}\log(0.1[OH^-]^3)\right) $
$ E_{\text{cell}}=0.46 -0.059\log[OH^-]+\frac{0.059}{2}\left(\log 0.1 + 3\log[OH^-]\right) $
Since $\log 0.1=-1$:
$ E_{\text{cell}}=0.46+\frac{0.059}{2}(-1) +\left(-0.059+\frac{3\cdot 0.059}{2}\right)\log[OH^-] $
$ E_{\text{cell}}=0.4305+0.0295\log[OH^-] $
Given $E_{\text{cell}}=0.2353\text{ V}$:
$ 0.2353=0.4305+0.0295\log[OH^-] \Rightarrow \log[OH^-]=\frac{-0.1952}{0.0295}\approx -6.6 $
So $pOH\approx 6.6 \Rightarrow pH\approx 14-6.6=7.4$.
Buffer calculation (H$_2$CO$_3$/HCO$_3^-$)
Henderson–Hasselbalch:
$ pH=pK_a+\log\frac{[HCO_3^-]}{[H_2CO_3]} $
$ 7.4=6.11+\log\frac{[HCO_3^-]}{[H_2CO_3]} \Rightarrow \log\frac{[HCO_3^-]}{[H_2CO_3]}=1.29 \Rightarrow \frac{[HCO_3^-]}{[H_2CO_3]}=19.5 $
Moles mixed:
$n(H_2CO_3)=2\,\text{M}\times 10\,\text{mL}=2\times 0.010=0.020\ \text{mol}$
$n(HCO_3^-)=5\,\text{M}\times x\,\text{mL}=5\times \frac{x}{1000}=0.005x\ \text{mol}$
Ratio $=19.5$:
$ \frac{0.005x}{0.020}=19.5 \Rightarrow 0.005x=0.39 \Rightarrow x=78 $
Answer: $\boxed{78\ \text{mL}}$
For strong electrolyte $\Lambda_m$ increases slowly with dilution and can be represented by the equation
$\Lambda_m = \Lambda_m^\circ - A c^{1/2}$
Molar conductivity values of the solutions of strong electrolyte AB at 18°C are given below :
| c [mol L-1] | 0.04 | 0.09 | 0.16 | 0.25 |
|---|---|---|---|---|
| $\Lambda_m$ [S cm2 mol-1] | 96.1 | 95.7 | 95.3 | 94.9 |
The value of constant A based on the above data [in S cm2 mol-1/(mol/L)1/2] unit is ________.
Explanation:
We use the given relation for a strong electrolyte:
Using equation : $\Lambda_{\mathrm{m}}=\Lambda_{\mathrm{m}}^0-\mathrm{A} \sqrt{\mathrm{c}}$
Now substitute the first set of data: $c = 0.04$ and $\Lambda_m = 96.1$.
$ \begin{aligned} & 96.1=\Lambda_{\mathrm{m}}^0-\mathrm{A} \sqrt{0.04} \\ & 96.1=\Lambda_{\mathrm{m}}^0-\mathrm{A} \times 0.2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & 95.7=\Lambda_{\mathrm{m}}^0-\mathrm{A} \times \sqrt{0.09} \\ & 95.7=\Lambda_{\mathrm{m}}^0-\mathrm{A} \times 0.3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $
Now subtract equation (ii) from equation (i) so that $\Lambda_{\mathrm{m}}^0$ cancels out:
From eq. (1) and eq. (2)
$ \begin{aligned} (96.1 - 95.7) &= (\Lambda_{\mathrm{m}}^0 - A \times 0.2) - (\Lambda_{\mathrm{m}}^0 - A \times 0.3) \\ 0.4 &= -0.2A + 0.3A \\ 0.4 &= 0.1A \\ A &= 4 \end{aligned} $
Consider the following redox reaction taking place in acidic medium
$ \mathrm{BH}_4^{-}(a q)+\mathrm{ClO}_3^{-}(a q) \longrightarrow \mathrm{H}_2 \mathrm{BO}_3^{-}(a q)+\mathrm{Cl}^{-}(a q) $
If the Nernst equation for the above balanced reaction is
$ \mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln \mathrm{Q}, $
then the value of $n$ is $\_\_\_\_$ .(Nearest integer)
Explanation:
Balance by ion–electron method in acidic medium.
Oxidation half-reaction (borohydride to dihydrogen borate):
Start: $\mathrm{BH_4^- \rightarrow H_2BO_3^-}$
Balance O by adding water:
$\mathrm{BH_4^- + 3H_2O \rightarrow H_2BO_3^-}$
Balance H by adding $\mathrm{H^+}$:
Left H = $4+6=10$, right H = $2$, so add $8\mathrm{H^+}$ on RHS:
$\mathrm{BH_4^- + 3H_2O \rightarrow H_2BO_3^- + 8H^+}$
Balance charge by adding electrons (oxidation ⇒ $e^-$ on RHS):
LHS charge $=-1$, RHS charge $=-1+8=+7$, so add $8e^-$ to RHS:
$\mathrm{BH_4^- + 3H_2O \rightarrow H_2BO_3^- + 8H^+ + 8e^-}$
So, electrons lost $=8$.
Reduction half-reaction (chlorate to chloride):
$\mathrm{ClO_3^- \rightarrow Cl^-}$
Balance O by adding water:
$\mathrm{ClO_3^- \rightarrow Cl^- + 3H_2O}$
Balance H by adding $\mathrm{H^+}$ on LHS:
$\mathrm{6H^+ + ClO_3^- \rightarrow Cl^- + 3H_2O}$
Balance charge by adding electrons (reduction ⇒ $e^-$ on LHS):
LHS charge $=6-1=+5$, RHS charge $=-1$, so add $6e^-$ to LHS:
$\mathrm{6e^- + 6H^+ + ClO_3^- \rightarrow Cl^- + 3H_2O}$
So, electrons gained $=6$.
Make electrons equal (LCM of $8$ and $6$ is $24$):
Multiply oxidation half by $3$ and reduction half by $4$:
Oxidation $\times 3$ gives $24e^-$, reduction $\times 4$ consumes $24e^-$.
After adding and canceling common species, the balanced overall reaction becomes:
$\mathrm{3BH_4^- + 4ClO_3^- \rightarrow 3H_2BO_3^- + 4Cl^- + 3H_2O}$
Hence, the number of electrons transferred in the balanced reaction is
$n = 24.$
So, $n = 24$ (nearest integer).
Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be $1 / 30$ of the molar conductivity of another weak acid HZ with concentration of 0.02 M . If $\lambda^{\circ} \mathrm{Q}^{-}$happened to be equal with $\lambda^{\circ} \mathrm{Z}^{-}$, then the difference of the $\mathrm{pK}_{\mathrm{a}}$ values of the two weak acids $\left(\mathrm{pK}_{\mathrm{a}}(\mathrm{HQ})-\mathrm{pK}_{\mathrm{a}}(\mathrm{HZ})\right)$ is $\_\_\_\_$ (Nearest integer).
[Given: degree of dissociation $(\alpha) \ll 1$ for both weak acids, $\lambda^{\circ}$ : limiting molar conductivity of ions]
Explanation:
$ \begin{array}{ll} \mathrm{K}_{\mathrm{a}}(\mathrm{HQ})=\mathrm{C}_1 \alpha_1^2 & \alpha_1=\frac{\lambda_{\mathrm{m}}(\mathrm{HQ})}{\lambda_{\mathrm{m}}^{\infty}(\mathrm{HQ})} \\ \mathrm{K}_{\mathrm{a}}(\mathrm{HZ})=\mathrm{C}_2 \alpha_2^2 & \alpha_2=\frac{\lambda_{\mathrm{m}}(\mathrm{HZ})}{\lambda_{\mathrm{m}}^{\infty}(\mathrm{HZ})} \end{array} $
$ \begin{aligned} & \frac{\mathrm{K}_{\mathrm{a}}(\mathrm{HQ})}{\mathrm{K}_{\mathrm{a}}(\mathrm{HZ})}=\frac{\mathrm{C}_1}{\mathrm{C}_2} \cdot\left(\frac{\alpha_1}{\alpha_2}\right)^2=\frac{0.18}{0.02} \cdot\left[\frac{\lambda_{\mathrm{m}}(\mathrm{HQ})}{\lambda_{\mathrm{m}}(\mathrm{HZ})}\right]^2 \\ & \frac{\mathrm{~K}_{\mathrm{a}}(\mathrm{HQ})}{\mathrm{K}_{\mathrm{a}}(\mathrm{HQ})}=9 \times\left(\frac{1}{30}\right)^2=\frac{1}{100} \\ & \mathrm{pK}_{\mathrm{a}}(\mathrm{HQ})-\mathrm{pK}_{\mathrm{a}}(\mathrm{HZ})=2 \end{aligned} $
Electricity is passed through an acidic solution of $\mathrm{Cu}^{2+}$ till all the $\mathrm{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL . The total volume of oxygen evolved at STP during the entire process is $\_\_\_\_$ mL . (Nearest integer)
[Given:
$ \begin{aligned} & \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{~s}) \mathrm{E}_{\mathrm{red}}^{\mathrm{o}}=+0.34 \mathrm{~V} \\ & \mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O} \mathrm{E}_{\mathrm{red}}^{\mathrm{o}}=+1.23 \mathrm{~V} \end{aligned} $
Molar mass of $\mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar mass of $\mathrm{O}_2=32 \mathrm{~g} \mathrm{~mol}^{-1}$
Faraday Constant $=96500 \mathrm{C} \mathrm{mol}^{-1}$
Molar volume at $\mathrm{STP}=22.4 \mathrm{~L}$ ]
Explanation:
Eq of $\mathrm{Cu}=\mathrm{Eq}$ of $\mathrm{O}_2$
In electrolysis, the number of equivalents deposited and liberated are equal for the same quantity of electricity.
For copper deposition: $\mathrm{Cu}^{2+}+2e^- \rightarrow \mathrm{Cu}$, so $n$-factor $=2$.
For oxygen evolution (reverse of given reduction): $2\mathrm{H}_2\mathrm{O}\rightarrow \mathrm{O}_2+4\mathrm{H}^+ +4e^-$, so $n$-factor $=4$.
So, equivalents of Cu deposited $=$ equivalents of $\mathrm{O}_2$ evolved during the first part.
$ \frac{300 \times 10^{-3} \times 2}{63.54}=\mathrm{n}_{\mathrm{O}_2} \times 4 $
$ 2.36 \times 10^{-3}=\mathrm{n}_{\mathrm{O}_2} $
When current is further passed
After all $\mathrm{Cu}^{2+}$ is exhausted, the current now causes only water oxidation, producing $\mathrm{O}_2$ at the anode.
Charge passed in 28 minutes at 600 mA:
$ \begin{aligned} & \mathrm{n}_{\mathrm{O}_2} \times 4=\frac{600 \times 28 \times 60}{96500 \times 1000} \\\\ & \mathrm{n}_{\mathrm{O}_2}=2.611 \times 10^{-3} \end{aligned} $
Total $\mathrm{O}_2$ released
Total moles of oxygen $=(2.36+2.611)\times 10^{-3}$.
Volume at STP $=n \times 22400\ \mathrm{mL}$.
$ \begin{aligned} & =\left[10^{-3} \times(2.36+2.611)\right] \times 22400 \mathrm{ml} \\\\ & =111.35 \mathrm{ml} \end{aligned} $
Consider the following electrochemical cell :
$ \mathrm{Pt}\left|\mathrm{O}_2(\mathrm{~g})(1 \mathrm{bar})\right| \mathrm{HCl}(\mathrm{aq}) \| \mathrm{M}^{2+}(\mathrm{aq}, 1.0 \mathrm{M}) \mid \mathrm{M}(\mathrm{~s}) $
The pH above which, oxygen gas would start to evolve at anode is $\_\_\_\_$ (nearest integer).
$ \left.\left[\begin{array}{ll} \text { Given : } & \mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^{\mathrm{o}}=0.994 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\mathrm{o}}=1.23 \mathrm{~V} \end{array}\right\} \text { standard reduction potential } \\ \text {and} \frac{\mathrm{RT}}{\mathrm{F}}(2.303)=0.059 \mathrm{~V} \text {at the given condition}\right] $
Explanation:
For the oxygen electrode (reduction reaction):
$\mathrm{O_2(g)}+4\mathrm{H^+ (aq)}+4e^- \rightarrow 2\mathrm{H_2O(l)}$
Nernst equation:
$ E_{\mathrm{O_2/H_2O}} = E^\circ_{\mathrm{O_2/H_2O}} - \frac{0.059}{4}\log\left(\frac{1}{p_{\mathrm{O_2}}[\mathrm{H^+}]^4}\right) $
Given $p_{\mathrm{O_2}}=1$ bar, so $\log p_{\mathrm{O_2}}=0$:
$ E_{\mathrm{O_2/H_2O}} = 1.23 + 0.059\log[\mathrm{H^+}] $
Since $\log[\mathrm{H^+}] = -\mathrm{pH}$:
$ E_{\mathrm{O_2/H_2O}} = 1.23 - 0.059\,\mathrm{pH} $
For the metal electrode:
$\mathrm{M^{2+}} + 2e^- \rightarrow \mathrm{M(s)}$
Given $[\mathrm{M^{2+}}]=1.0\ \mathrm{M}$, hence
$ E_{\mathrm{M^{2+}/M}} = E^\circ_{\mathrm{M^{2+}/M}} = 0.994\ \mathrm{V} $
Oxygen will start evolving at the anode when the oxygen electrode becomes the anode, i.e. when its reduction potential becomes less than that of $\mathrm{M^{2+}/M}$. The threshold is when both are equal:
$ 1.23 - 0.059\,\mathrm{pH} = 0.994 $
$ 0.059\,\mathrm{pH} = 1.23 - 0.994 = 0.236 $
$ \mathrm{pH} = \frac{0.236}{0.059} = 4 $
So, oxygen gas starts evolving at the anode for $\mathrm{pH} > 4$.
$ \boxed{4} $
Consider the following electrochemical cell at 298 K
$\mathrm{Pt}\left|\mathrm{HSnO}_2^{-}(\mathrm{aq})\right| \mathrm{Sn}(\mathrm{OH})_6{ }^{2-}(\mathrm{aq})\left|\mathrm{OH}^{-}(\mathrm{aq})\right| \mathrm{Bi}_2 \mathrm{O}_3(\mathrm{~s}) \mid \mathrm{Bi}(\mathrm{s})$.
If the reaction quotient at a given time is $10^6$, then the cell EMF $\left(\mathrm{E}_{\text {cell }}\right)$ is
$\_\_\_\_$ $\times 10^{-1} \mathrm{~V}$ (Nearest integer).
Given the standard half-cell reduction potential as
$ \mathrm{E}_{\mathrm{Bi}_2 \mathrm{O}_3 / \mathrm{Bi}, \mathrm{OH}^{-}}^{\circ}=-0.44 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{Sn}(\mathrm{OH})_6^{2-} / \mathrm{HSnO}_2^{-}, \mathrm{OH}^{-}}^{\circ}=-0.90 \mathrm{~V} $
Explanation:
$ \begin{aligned} \mathrm{E}_{\mathrm{cell}}^{\circ} & =-0.44-(-0.90) \\ & =+0.46 \mathrm{~V} \end{aligned} $
Applying Nernst equation :-
$ \begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.06}{\mathrm{n}} \log \mathrm{Q} \\ & \text{First, find } n \text{ (number of electrons transferred). In this cell, } n=6. \\ & \text{Given } \mathrm{Q}=10^6 \text{, so } \log 10^6=6. \\ & \Rightarrow \mathrm{E}_{\text {cell }}=0.46-\frac{0.06}{6} \log 10^6 \\ & \Rightarrow \mathrm{E}_{\text {cell }}=0.46-\frac{0.06}{6}\times 6 \\ & \Rightarrow \mathrm{E}_{\text {cell }}=0.46-0.06 \\ & \Rightarrow \mathrm{E}_{\text {cell }}=0.40 \mathrm{~V}=4 \times 10^{-1} \mathrm{~V \\ } \\& \Rightarrow \mathrm{x}=4 \end{aligned} $
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
$\mathrm{MX}(\mathrm{s}) \rightleftharpoons \mathrm{M}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}) ; \quad \mathrm{K}_{\mathrm{sp}}=10^{-10}$
If the standard reduction potential for M+ (aq) + e− → M(s) is
$\left(\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}^{\ominus}\right)=0.79 \mathrm{~V}$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $\mathrm{E}_{\mathrm{X}^{-} / \mathrm{MX}(\mathrm{s}) / \mathrm{M}}^{\ominus}$ is ______ mV. (nearest integer)
[Given: $ \dfrac{2.303 RT}{F} = 0.059\ \text{V} $]
Explanation:
For the metal/metal insoluble salt electrode, the reduction half-reaction is
$\mathrm{MX(s)}+e^- \rightarrow \mathrm{M(s)}+\mathrm{X^- (aq)}$
We can obtain this by adding the following two steps:
1) Solubility equilibrium:
$\mathrm{MX(s)} \rightleftharpoons \mathrm{M^+(aq)}+\mathrm{X^-(aq)},\quad K_{sp}=10^{-10}$
2) Reduction of $\mathrm{M^+}$:
$\mathrm{M^+(aq)}+e^- \rightarrow \mathrm{M(s)},\quad E^\circ =0.79\ \text{V}$
Adding (1) and (2), $\mathrm{M^+}$ cancels and we get the required electrode reaction.
Now use $\Delta G^\circ$ relation (NCERT):
For equilibrium: $\Delta G^\circ = -RT\ln K$
For a half-cell: $\Delta G^\circ = -nFE^\circ$
So,
For dissolution:
$ \Delta G^\circ_{\text{diss}}=-RT\ln(10^{-10}) = -RT(-10\ln 10)= 10RT\ln 10 $
For reduction of $\mathrm{M^+}$:
$ \Delta G^\circ_{\mathrm{M^+/M}}=-F(0.79) $
Total:
$ \Delta G^\circ_{\text{total}}=10RT\ln 10 - F(0.79) $
For the overall electrode reaction ($n=1$):
$ E^\circ_{\mathrm{X^- /MX(s)/M}} = -\frac{\Delta G^\circ_{\text{total}}}{F} = -\left(\frac{10RT\ln 10}{F}-0.79\right) $
Given $\dfrac{2.303RT}{F}=0.059\ \text{V}$ and $\ln 10=2.303$,
$ \frac{RT\ln 10}{F}=\frac{2.303RT}{F}=0.059\ \text{V} $
$ \frac{10RT\ln 10}{F}=10\times 0.059=0.59\ \text{V} $
So,
$ E^\circ = -\left(0.59-0.79\right)=0.20\ \text{V}=200\ \text{mV} $
Answer: $200\ \text{mV}$
The pH and conductance of a weak acid $(\mathrm{HX})$ was found to be 5 and $4 \times 10^{-5} \mathrm{~S}$, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of $1 \mathrm{~cm}^2$ were at a distance of 15 cm apart. The value of the limiting molar conductivity is $\_\_\_\_$ $\mathrm{S} \mathrm{m}^2 \mathrm{~mol}^{-1}$. (nearest integer)
(Given : degree of dissociation of the weak acid $(\alpha) \ll 1$ )
Explanation:
Step 1: Find $[{H}^+]$ from pH
The pH is given as 5.
So, $[{\mathrm{H}}^+] = 10^{-5}~\mathrm{mol~L}^{-1}$
Step 2: Write the relation for weak acid dissociation
For a weak acid $HX$:
$[{\mathrm{H}}^+] = [HX] \cdot \alpha$
Where $[HX]$ is the initial concentration and $\alpha$ is the degree of dissociation.
Step 3: Link $\alpha$ and molar conductivity
Degree of dissociation is also given by
$\alpha = \dfrac{\Lambda_m}{\Lambda_m^\infty}$
Where $\Lambda_m$ is the molar conductivity at given concentration, and $\Lambda_m^\infty$ is the molar conductivity at infinite dilution (limiting molar conductivity).
Step 4: Find molar conductivity $\Lambda_m$
Molar conductivity is given by:
$\Lambda_m = \dfrac{k \times 1000}{[HX]}$
$k$ is the specific conductance (conductivity).
Step 5: Calculate cell constant and specific conductance ($k$)
Given conductance (G) $= 4 \times 10^{-5}~\mathrm{S}$
Cell constant $= \dfrac{\text{distance between plates}}{\text{area of plates}} = \dfrac{15~\mathrm{cm}}{1~\mathrm{cm}^2} = 15~\mathrm{cm}^{-1}$
So, $k = G \times \text{cell constant}$
$\rightarrow k = 4 \times 10^{-5} \times 15 = 6 \times 10^{-4}~\mathrm{S~cm}^{-1}$
Step 6: Combine equations to solve for $\Lambda_m^\infty$
From above, $[{\mathrm{H}}^+] = [HX] \cdot \alpha$
But, $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\infty}$, so:
$[{\mathrm{H}}^+] = [HX] \cdot \dfrac{\Lambda_m}{\Lambda_m^\infty}$
Recall $\Lambda_m = \dfrac{k \times 1000}{[HX]}$
Substitute in:
$[{\mathrm{H}}^+] = [HX] \cdot \dfrac{\dfrac{k \times 1000}{[HX]}}{\Lambda_m^\infty}$
$ = \dfrac{k \times 1000}{\Lambda_m^\infty}$
So, $\Lambda_m^\infty = \dfrac{k \times 1000}{[{\mathrm{H}}^+]}$
Step 7: Substitute values and solve
$k = 6 \times 10^{-4}~\mathrm{S~cm}^{-1}$
$[{\mathrm{H}}^+] = 10^{-5}~\mathrm{mol~L}^{-1}$
So,
$ \Lambda_m^\infty = \frac{6 \times 10^{-4} \times 1000}{10^{-5}} = \frac{6 \times 10^{-1}}{10^{-5}} = 6 \times 10^{4}~\mathrm{S~cm}^2\mathrm{~mol}^{-1} $
Step 8: Convert units to $\mathrm{S~m}^2\mathrm{~mol}^{-1}$
$1~\mathrm{S~cm}^2\mathrm{~mol}^{-1} = 10^{-4}~\mathrm{S~m}^2\mathrm{~mol}^{-1}$
So, $ 6 \times 10^{4} \times 10^{-4} = 6~\mathrm{S~m}^2 \mathrm{~mol}^{-1} $
Final Answer:
Limiting molar conductivity $\Lambda_m^\infty = 6~\mathrm{S~m}^2 \mathrm{~mol}^{-1}$.
Consider the above electrochemical cell where a metal electrode ( M ) is undergoing redox reaction by forming $\mathrm{M}^{+}\left(\mathrm{M} \rightarrow \mathrm{M}^{+}+\mathrm{e}^{-}\right)$. The cation $\mathrm{M}^{+}$is present in two different concentrations $c_1$ and $c_2$ as shown above. Which of the following statement is correct for generating a positive cell potential?
If $c_1$ is present at anode, then $c_1>c_2$.
If $c_1$ is present at cathode, then $c_1>c_2$.
If $c_1$ is present at cathode, then $c_1
If $c_1$ is present at anode, then $c_1=c_2$.
In the given electrochemical cell, $\mathrm{Ag}(\mathrm{s})|\mathrm{AgCl}(\mathrm{s})| \mathrm{FeCl}_2(\mathrm{aq}), \mathrm{FeCl}_3(\mathrm{aq}) \mid \mathrm{Pt}(\mathrm{s})$ at 298 K , the cell potential ( $\mathrm{E}_{\text {cell }}$ ) will increase when :
A. Concentration of $\mathrm{Fe}^{2+}$ is increased.
B. Concentration of $\mathrm{Fe}^{3+}$ is decreased.
C. Concentration of $\mathrm{Fe}^{2+}$ is decreased.
D. Concentration of $\mathrm{Fe}^{3+}$ is increased.
E. Concentration of $\mathrm{Cl}^{-}$is increased.
Choose the correct answer from the options given below :
C, D and E Only
A and B Only
B Only
A and E Only
Consider the following reduction processes :
$ \begin{aligned} & \mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(\mathrm{~s}), \mathrm{E}^0=-1.66 \mathrm{~V} \\ & \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}, \mathrm{E}^0=+0.77 \mathrm{~V} \\ & \mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+}, \mathrm{E}^0=+1.81 \mathrm{~V} \\ & \mathrm{Cr}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Cr}(\mathrm{~s}), \mathrm{E}^0=-0.74 \mathrm{~V} \end{aligned} $
The tendency to act as reducing agent decreases in the order :
$\mathrm{Al}>\mathrm{Fe}^{2+}>\mathrm{Cr}>\mathrm{Co}^{2+}$
$\mathrm{Al}>\mathrm{Cr}>\mathrm{Co}^{2+}>\mathrm{Fe}^{2+}$
$\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Al}>\mathrm{Co}^{2+}$
$\mathrm{Al}>\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Co}^{2+}$
Given at 298 K :
$ \begin{aligned} & \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\ominus}=\mathrm{X} \text { Volt } \\ & \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}}^{\ominus}=\mathrm{Y} \text { Volt } \end{aligned} $
The $\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\ominus}$ in Volt at 298 K is given by :
$2 \mathrm{X}-3 \mathrm{Y}$
$3 Y-2 X$
$3 \mathrm{Y}+2 \mathrm{X}$
$Y+X$
For a general redox reaction
Anode $\quad \operatorname{Red}_1 \rightarrow \mathrm{Ox}_1^{\mathrm{n}_1^{+}}+\mathrm{n}_1 \mathrm{e}^{-}$
Cathode $\mathrm{O} x_2+\mathrm{n}_2 \mathrm{e}^{-} \rightarrow \operatorname{Red}_2^{\mathrm{n}_2-}$
Which of the following statement is incorrect?
$ \begin{aligned} & \text { The overall reaction can be written as } \mathrm{n}_2 \operatorname{Red}_1+\mathrm{n}_1 \mathrm{O} x_2 \rightleftharpoons \mathrm{n}_2 \mathrm{O} x_1^{\mathrm{n}_1^{+}}+\mathrm{n}_1 \operatorname{Red}_2^{\mathrm{n}_2^{-}} \\ & \text {} \end{aligned} $
The electrons do not appear in the overall reaction because electrons produced at the anode are consumed at the cathode.
Here $n$ is the number of electrons transferred in redox reaction.
If the reaction is carried out reversibly, the electrical work done is equal to the ratio of charge and potential difference through which charge is moved.
$ \text { Consider the following data. } $
$ \begin{array}{|c|c|} \hline \text { Electrolyte } & \wedge^{\circ}_\mathbf{m}{\mathbf{(}} \mathbf{S ~ c m}^{\mathbf{2}} \mathbf{~ m o l}^{\mathbf{1}} \mathbf{)} \\ \hline \mathrm{BaCl}_2 & x_1 \\ \hline \mathrm{H}_2 \mathrm{SO}_4 & x_2 \\ \hline \mathrm{HCl} & x_3 \\ \hline \end{array} $
$\mathrm{BaSO}_4$ is sparingly soluble in water. If the conductivity of the saturated $\mathrm{BaSO}_4$ solution is $x \mathrm{~S} \mathrm{~cm}^{-1}$ then the solubility product of $\mathrm{BaSO}_4$ can be given as (Here $\wedge_{\mathrm{m}}=\wedge^{\circ}_{\mathrm{m}}$ )
$\frac{10^6 x^2}{\alpha^2\left(x_1+x_2-2 x_3\right)^2}$
$ \frac{x^2}{\left(x_1+x_2-2 x_3\right)^2} $
$ \frac{\alpha^2\left(x_1+x_2-2 x_3\right)^2}{10^6 x^2} $
$ \frac{x^2}{\left(x_1+x_2+2 x_3\right)^2} $
One half cell in a voltaic cell is constructed by dipping silver rod in $\mathrm{AgNO}_3$ solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of $\mathrm{ZnSO}_4$.
A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of $\mathrm{Ag}^{+}$ions used in terms of $\log x\left(x=\left[\mathrm{Ag}^{+}\right]\right)$?
$ \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\ominus}=-0.76 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\ominus}=+0.80 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.059 \mathrm{~V} $
$\frac{2}{3.9}$
$\frac{4}{5.9}$
$\frac{2.9}{2}$
$\frac{5.9}{4}$
An electrochemical cell is constructed using half cells in the direction of spontaneous change:
Fe(OH)2(s) + 2e− → Fe(s) + 2OH−(aq) Eθ = −0.88 V
and AgBr(s) + e− → Ag(s) + Br−(aq) Eθ = +0.07 V
Which of the following option is correct?
$E_{cell}^{\theta} = -0.95 \ V$
Fe is reduced in the electrochemical cell
$E_{cell}^{\theta}$ is an extensive property
At 298 K , the molar conductivity of $x \%(\mathrm{w} / \mathrm{w}) \mathrm{MX}$ solution (aqueous) is $123.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$. The conductance of same solution is $1.9 \times 10^{-3} \mathrm{~S}$. The value of $x$ is $\_\_\_\_$ $\times 10^{-2}$.
(Given: cell constant $=1.3 \mathrm{~cm}^{-1}$; molar mass of MX is $75 \mathrm{~g} \mathrm{~mol}^{-1}$, density of aqueous solution of MX at 298 K is $1.0 \mathrm{~g} \mathrm{~mL}^{-1}$ )
Explanation:
Conductivity ($\kappa$) is related to the conductance ($G$) and the cell constant ($G^*$) by the following relationship:
$ \kappa = G \times G^* $
Substituting the given values:
$ \kappa = (1.9 \times 10^{-3} \mathrm{~S}) \times (1.3 \mathrm{~cm}^{-1}) $
$ \kappa = 2.47 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1} $
The molar conductivity ($\Lambda_m$) is related to the conductivity ($\kappa$) and the molarity ($C$ in $\mathrm{mol~L}^{-1}$) by the formula:
$ \Lambda_m = \frac{\kappa \times 1000}{C} $
Rearranging the formula to solve for molarity ($C$):
$ C = \frac{\kappa \times 1000}{\Lambda_m} $
Substituting the values into the equation:
$ C = \frac{(2.47 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1}) \times 1000 \mathrm{~cm}^3 \mathrm{~L}^{-1}}{123.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}} $
$ C = \frac{2.47}{123.5} \mathrm{~mol \mathrm{~L}^{-1}} $
$ C = 0.02 \mathrm{~mol \mathrm{~L}^{-1}} $
An $x \% (\mathrm{w}/\mathrm{w})$ aqueous solution means there are $x \mathrm{~g}$ of the solute (MX) in $100 \mathrm{~g}$ of the solution.
Using the density of the solution, we can find the volume of $100 \mathrm{~g}$ of the solution:
$ \text{Volume of solution} = \frac{\text{Mass of solution}}{\text{Density}} $
$ \text{Volume} = \frac{100 \mathrm{~g}}{1.0 \mathrm{~g} \mathrm{~mL}^{-1}} = 100 \mathrm{~mL} $
Molarity is defined as the number of moles of solute per $1000 \mathrm{~mL}$ (1 liter) of the solution.
Since $100 \mathrm{~mL}$ of the solution contains $x \mathrm{~g}$ of MX, $1000 \mathrm{~mL}$ of the solution will contain $10x \mathrm{~g}$ of MX.
Now, calculate the number of moles of MX in $1000 \mathrm{~mL}$, which is exactly our molarity ($C$):
$ \text{Moles in 1 L} = \frac{\text{Mass of solute in 1 L}}{\text{Molar mass of solute}} $
$ C = \frac{10x}{75} $
We already calculated the molarity to be $0.02 \mathrm{~mol \mathrm{~L}^{-1}}$. Equating the two values:
$ \frac{10x}{75} = 0.02 $
$ 10x = 0.02 \times 75 $
$ 10x = 1.5 $
$ x = 0.15 $
The problem requires $x$ to be expressed in the form of $\_\_\_\_ \times 10^{-2}$.
$ x = 0.15 = 15 \times 10^{-2} $
The value of $x$ is 15.
An electrochemical cell, consist of the following two redox couples, $\mathrm{M}^{x+} (\mathrm{aq}) / \mathrm{M}(\mathrm{s})\left[\mathrm{E}_{\text {red }}^{\Theta}=+0.15 \mathrm{~V}\right]$ and $\mathrm{Fe}^{3+}(\mathrm{aq}) / \mathrm{Fe}(\mathrm{s})\left[\mathrm{E}_{\text {red }}^{\Theta}=-0.036 \mathrm{~V}\right]$. The cell EMF $\left(\mathrm{E}_{\text {cell }}\right)$ is recorded to be 0.2057 V . If the reaction quotient of the electrochemical reaction is found to be $10^{-2}$, then the value of $x$ is
$\_\_\_\_$ .(Nearest integer)
[Given : M is a p-block metal and $\frac{2.303 R T}{F}=0.059 \mathrm{~V}$ ]
Explanation:
For the cell, the two reduction potentials are:
$ \mathrm{M}^{x+} + x e^- \rightarrow \mathrm{M}(s), \qquad E^\circ_{\text{red}} = +0.15\,\text{V} $
and
$ \mathrm{Fe}^{3+} + 3 e^- \rightarrow \mathrm{Fe}(s), \qquad E^\circ_{\text{red}} = -0.036\,\text{V} $
Since the reduction potential of $\mathrm{M}^{x+}/\mathrm{M}$ is greater, it will act as the cathode.
So,
Cathode: $\mathrm{M}^{x+} + x e^- \rightarrow \mathrm{M}(s)$
Anode: $\mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{3+} + 3e^-$
Hence,
$ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.15 - (-0.036) = 0.186\,\text{V} $
Now use the Nernst equation:
$ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log Q $
Given:
$ E_{\text{cell}} = 0.2057\,\text{V}, \qquad Q = 10^{-2} $
Since $\log 10^{-2} = -2$,
$ 0.2057 = 0.186 - \frac{0.059}{n}(-2) $
$ 0.2057 = 0.186 + \frac{0.118}{n} $
$ 0.2057 - 0.186 = \frac{0.118}{n} $
$ 0.0197 = \frac{0.118}{n} $
$ n = \frac{0.118}{0.0197} \approx 5.99 \approx 6 $
So the number of electrons transferred in the balanced cell reaction is
$ n = 6 $
Now, for balancing electrons:
Oxidation of iron gives $3$ electrons
Reduction of $\mathrm{M}^{x+}$ consumes $x$ electrons
Therefore, total electrons transferred in the balanced equation will be $\mathrm{LCM}(3,x)$.
Given this is $6$, we get:
$ \mathrm{LCM}(3,x)=6 $
Possible values of $x$ are $2$ or $6$.
Now it is given that $\mathrm{M}$ is a p-block metal. Common p-block metals generally form ions like $2^+$, $3^+$, $4^+$, etc. Among the possible values, the most suitable is
$ x=2 $
So, the nearest integer value of $x$ is
$ \boxed{2} $
Consider the following two half-cell reactions along with the standard reduction potential given :
$\mathrm{CO}_2 + 6\mathrm{H}^+ + 6e^- \longrightarrow \mathrm{CH}_3\mathrm{OH} + \mathrm{H}_2\mathrm{O}\qquad E_{\text{red}}^{\circ} = 0.02\ \mathrm{V}$
$\dfrac{1}{2} \mathrm{O}_2 + 2\mathrm{H}^+ + 2e^- \longrightarrow \mathrm{H}_2\mathrm{O}\qquad E_{\text{red}}^{\circ} = 1.23\ \mathrm{V}$
A fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 bar pressure and 298 K temperature. The fuel cell works with 80% efficiency. If the work derived from the cell using 1 mol of $\mathrm{CH}_3\mathrm{OH}$ is used to compress an ideal gas isothermally against a constant pressure of 1 kPa, then the change in the volume of the gas,
$\Delta V = \underline{\hspace{2cm}}$ m$^3$. (nearest integer)
Given : $\mathrm{F} = 96500~\mathrm{C~mol}^{-1}$
Explanation:
For the fuel cell, methanol will be oxidised and oxygen will be reduced.
Given reduction half-cells:
$ \mathrm{CO}_2 + 6\mathrm{H}^+ + 6e^- \longrightarrow \mathrm{CH}_3\mathrm{OH} + \mathrm{H}_2\mathrm{O} \qquad E^\circ_{\text{red}} = 0.02\ \mathrm{V} $
$ \frac{1}{2}\mathrm{O}_2 + 2\mathrm{H}^+ + 2e^- \longrightarrow \mathrm{H}_2\mathrm{O} \qquad E^\circ_{\text{red}} = 1.23\ \mathrm{V} $
Since methanol is the fuel, it will undergo oxidation. So we reverse the first reaction:
$ \mathrm{CH}_3\mathrm{OH} + \mathrm{H}_2\mathrm{O} \longrightarrow \mathrm{CO}_2 + 6\mathrm{H}^+ + 6e^- $
Its oxidation potential is:
$ E^\circ_{\text{ox}} = -0.02\ \mathrm{V} $
Now calculate standard cell potential:
$ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} + E^\circ_{\text{anode(ox)}} $
$ E^\circ_{\text{cell}} = 1.23 + (-0.02) = 1.21\ \mathrm{V} $
Number of electrons transferred:
$ n = 6 $
So maximum electrical work from 1 mol of $\mathrm{CH}_3\mathrm{OH}$ is
$ w_{\max} = nFE^\circ_{\text{cell}} $
$ w_{\max} = 6 \times 96500 \times 1.21 $
$ w_{\max} = 700590\ \mathrm{J} $
Fuel cell works with $80\%$ efficiency, so actual work obtained is
$ w = 0.8 \times 700590 $
$ w = 560472\ \mathrm{J} $
This work is used to compress an ideal gas isothermally against constant pressure $P = 1\ \text{kPa} = 1000\ \text{Pa}$.
For compression against constant external pressure,
$ w = P\Delta V $
So,
$ \Delta V = \frac{w}{P} $
$ \Delta V = \frac{560472}{1000} = 560.472\ \mathrm{m}^3 $
Nearest integer:
$ \Delta V \approx 560\ \mathrm{m}^3 $
Since it is compression, volume decreases. So strictly,
$ \Delta V = -560\ \mathrm{m}^3 $
If only magnitude is asked, then
$ \boxed{560} $
Consider the following half cell reaction
$ \text{Cr}_2\text{O}_7^{2-} \, (\text{aq}) + 6\text{e}^- + 14\text{H}^+ \, (\text{aq}) \rightarrow 2\text{Cr}^{3+} \, (\text{aq}) + 7\text{H}_2\text{O} \, (\ell) $
The reaction was conducted with the ratio of $\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}$. The pH value at which the EMF of the half cell will become zero is __________.
(nearest integer value)
[Given: standard half cell reduction potential $E^{\circ}_{\text{Cr}_2\text{O}_7^{2-}, \text{H}^+/\text{Cr}^{3+}} = 1.33\, \text{V}$, $\frac{2.303RT}{F} = 0.059\, \text{V}$.]
Explanation:
$\begin{aligned} & \mathrm{Cr}_2 \mathrm{O}_{7(\mathrm{aq})}^{-2}+14 \mathrm{H}_{(\mathrm{aq})}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{+3}+7 \mathrm{H}_2 \mathrm{O}_{(\ell)} \\ & \mathrm{E}_{\mathrm{R}}=\mathrm{E}_{\mathrm{R}}^0-\frac{0.059}{6} \log \frac{\left[\mathrm{Cr}^{+3}\right]^2}{\left[\mathrm{Cr}_2 \mathrm{O}_7^{-2}\right]\left[\mathrm{H}^{+}\right]^{14}} \\ & 0=1.33-\frac{0.059}{6} \log \frac{10^{-6}}{\left[\mathrm{H}^{+}\right]^{14}} \\ & \frac{1.33 \times 6}{0.059}=\log \frac{10^{-6}}{[\mathrm{H}]^{14}} \\ & 135.254=-6-14 \log \left[\mathrm{H}^{+}\right] \\ & 141.254=14 \mathrm{pH} \\ & \mathrm{pH}=\frac{141.254}{14}=10.08 \end{aligned}$
1 Faraday electricity was passed through $\mathrm{Cu}^{2+}(1.5 \mathrm{M}, 1 \mathrm{~L}) / \mathrm{Cu}$ and 0.1 Faraday was passed through $\mathrm{Ag}^{+}(0.2 \mathrm{M}, 1 \mathrm{~L}) / \mathrm{Ag}$ electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is __________ mV (nearest integer)
$\begin{aligned} \text { Given : } \mathrm{E}^{\circ} \mathrm{Cu}^{2+} / \mathrm{Cu} & =0.34 \mathrm{~V} \\\\ \mathrm{E}^{\circ} \mathrm{Ag}^{+} / \mathrm{Ag} & =0.8 \mathrm{~V} \\\\ \frac{2 \cdot 303 \mathrm{RT}}{\mathrm{~F}} & =0.06 \mathrm{~V} \end{aligned}$
Explanation:
reaction
$\begin{aligned} & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.06}{\mathrm{n}} \log \frac{\left[\mathrm{Cu}^{+2}\right]}{\left[\mathrm{Ag}^{+}\right]^2} \\ & \mathrm{E}=(0.8-0.34)-\frac{0.06}{2} \log \frac{1}{(0.1)^2}=0.4 \mathrm{~V} \end{aligned}$
Correct answer $=400 \mathrm{mV}$
$0.2 \%(\mathrm{w} / \mathrm{v})$ solution of NaOH is measured to have resistivity $870.0 \mathrm{~m} \Omega \mathrm{~m}$. The molar conductivity of the solution will be__________$\times 10^2 \mathrm{mS} \mathrm{dm}^2 \mathrm{~mol}^{-1}$. (Nearest integer)
Explanation:
Determine the conductivity from the resistivity.
The resistivity is given as
$\rho = 870.0\; \text{m}\Omega\cdot\text{m}.$
Convert this into ohm·meters:
$870.0\; \text{m}\Omega\cdot\text{m} = 870.0 \times 10^{-3}\; \Omega\cdot\text{m} = 0.87\; \Omega\cdot\text{m}.$
Conductivity is the reciprocal of resistivity:
$\kappa = \frac{1}{\rho} = \frac{1}{0.87} \approx 1.15\; \text{S/m}.$
Find the concentration of NaOH in mol/m³.
A $0.2\%\;(\mathrm{w}/\mathrm{v})$ solution means there are 0.2 grams of NaOH per 100 mL of solution.
In 1 liter (1000 mL) there are:
$\frac{0.2\; \text{g}}{100\; \text{mL}} \times 1000\; \text{mL} = 2\; \text{g/L}.$
The molar mass of NaOH is approximately $40\; \text{g/mol}$. Thus, the molarity is:
$\text{Molarity} = \frac{2\; \text{g/L}}{40\; \text{g/mol}} = 0.05\; \text{mol/L}.$
Since $1\; \text{L} = 0.001\; \text{m}^3$, converting to SI units (mol/m³):
$c = 0.05\; \text{mol/L} \times \frac{1}{0.001\; \text{m}^3/\text{L}} = 50\; \text{mol/m}^3.$
Calculate the molar conductivity in SI units.
Molar conductivity $\Lambda_m$ is given by:
$\Lambda_m = \frac{\kappa}{c}.$
Substituting the values:
$\Lambda_m = \frac{1.15\; \text{S/m}}{50\; \text{mol/m}^3} = 0.023\; \text{S}\cdot \text{m}^2/\text{mol}.$
Convert the molar conductivity to the desired units (mS dm² mol⁻¹).
First, convert the conductivity:
$0.023\; \text{S}\cdot \text{m}^2/\text{mol} \times 1000\; \frac{\text{mS}}{\text{S}} = 23\; \text{mS}\cdot \text{m}^2/\text{mol}.$
Now, convert the area units. Recall that:
$1\; \text{m}^2 = 100\; \text{dm}^2.$
So,
$23\; \text{mS}\cdot \text{m}^2/\text{mol} = 23 \times 100\; \text{mS}\cdot \text{dm}^2/\text{mol} = 2300\; \text{mS}\cdot \text{dm}^2/\text{mol}.$
The problem asks for the answer in the form
$\text{[blank]} \times 10^2\; \text{mS}\cdot \text{dm}^2/\text{mol}.$
Expressing 2300 mS dm²/mol in this form:
$2300\; \text{mS}\cdot \text{dm}^2/\text{mol} = 23 \times 10^2\; \text{mS}\cdot \text{dm}^2/\text{mol}.$
Thus, the molar conductivity is
$\boxed{23 \times 10^2\; \text{mS}\cdot \text{dm}^2/\text{mol}}.$
Consider the following electrochemical cell at standard condition.
$\mathrm{Au}(\mathrm{~s})\left|\mathrm{QH}_2, \mathrm{Q}\right| \mathrm{NH}_4 \mathrm{X}(0.01 \mathrm{M})| | \mathrm{Ag}^{+}(1 \mathrm{M}) \mid \mathrm{Ag}(\mathrm{~s}) \mathrm{E}_{\text {cell }}=+0.4 \mathrm{~V}$
The couple $\mathrm{QH}_2 / \mathrm{Q}$ represents quinhydrone electrode, the half cell reaction is given below:
$\left[\text { Given : } \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=+0.8 \mathrm{~V} \text { and } \frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.06 \mathrm{~V}\right]$
The $\mathrm{pK}_{\mathrm{b}}$ value of the ammonium halide salt $\left(\mathrm{NH}_4 \mathrm{X}\right)$ used here is __________ . (nearest integer)
Explanation:
The cell reaction is:
$ \mathrm{QH}_2 + 2 \mathrm{Ag}^{+} \rightarrow 2 \mathrm{Ag} + \mathrm{Q} + 2 \mathrm{H}^{+} $
The Nernst equation for the reaction can be expressed as:
$ \mathrm{E} = \mathrm{E}^{\circ} - \frac{0.06}{2} \log \left[\mathrm{H}^{+}\right]^2 $
Simplifying, we obtain:
$ \mathrm{E} = \mathrm{E}^{\circ} - 0.06 \log \left[\mathrm{H}^{+}\right] $
Given data includes:
$\mathrm{E}_{\text{cell}} = +0.4 \, \mathrm{V}$
Standard potential for $\mathrm{Ag}^{+}/\mathrm{Ag}$, $\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0 = +0.8 \, \mathrm{V}$
Using the cell potential and the standard potential, we calculate:
$ \mathrm{pH} = -\log \left(\mathrm{H}^{+}\right) = \frac{\mathrm{E} - \mathrm{E}^{\circ}}{0.06} = \frac{0.4 - 0.1}{0.06} $
$ \mathrm{pH} = \frac{0.3}{0.06} = 5 $
This established pH must now relate to the buffer equation involving $\mathrm{NH}_4 \mathrm{X}$, at
$ \mathrm{pH} + \mathrm{NH}_4 \mathrm{X} = 7 - \frac{1}{2} \mathrm{pK}_{\mathrm{b}} - \frac{1}{2} \log \mathrm{C} $
Given the concentration $\mathrm{C} = 0.01 \, \mathrm{M} = 10^{-2}$, we substitute into the equation:
$ 5 = 7 - \frac{1}{2} \times \mathrm{pK}_{\mathrm{b}} - \frac{1}{2} \log (10^{-2}) $
Simplifying further, the equation resolves to:
$ \mathrm{pK}_{\mathrm{b}} = 6 $
Thus, the calculated $\mathrm{pK}_{\mathrm{b}}$ value of the ammonium halide salt $\mathrm{NH}_4 \mathrm{X}$ is 6.
The current in Amperes used for the given electrolysis is ___________ . (Nearest integer).
Explanation:
Electrolysis of NaCl is
$\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}(\mathrm{aq}) \rightarrow \mathrm{NaOH}(\mathrm{aq})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g})+\frac{1}{2} \mathrm{H}_2(\mathrm{~g})$
Since during electrolysis pH changes to 12
So $\left[\mathrm{OH}^{\ominus}\right]=10^{-2}$ and $\left[\mathrm{H}^{+}\right]=10^{-12}$
So by Faraday law
Gram amount of substance deposited $=$ Amount of electricity passed
$\begin{aligned} & 10^{-2} \times \frac{600}{1000} \times 96500=\mathrm{I} \times \mathrm{t} \\ & \frac{10^{-2} \times 600}{1000} \times 96500=\mathrm{I} \times 5 \times 60 \\ & \mathrm{I}=\frac{10^{-2} \times 600 \times 96500}{1000 \times 5 \times 60} \\ & \mathrm{I}=1.93 \text { ampere } \end{aligned}$
So, $\mathrm{I = 2}$ ampere (nearest integer)
Given below is the plot of the molar conductivity vs $\sqrt{\text { concentration }}$ for KCl in aqueous solution.
If, for the higher concentration of KCl solution, the resistance of the conductivity cell is $100 \Omega$, then the resistance of the same cell with the dilute solution is ' x ' $\Omega$
The value of $x$ is _________ (Nearest integer)
Explanation:
Resistance and Resistivity:
$ R = \rho \frac{\ell}{A} $
Where $ R $ is resistance, $ \rho $ is resistivity, $ \ell $ is the length, and $ A $ is the cross-sectional area of the cell. The term $ \frac{\ell}{A} $ is known as the cell constant, $ G^* $.
Conductivity and Conductance:
$ \kappa = G \cdot G^* $
$ G = \frac{1}{R} \quad ; \quad \kappa = \frac{1}{\rho} $
Where $ G $ is conductance and $ \kappa $ is conductivity.
Molar Conductivity:
$ \lambda_m = \frac{\kappa \times 1000}{C} $
Where $ \lambda_m $ is molar conductivity and $ C $ is concentration.
Given Information:
The resistance of the cell with the concentrated solution $ R_c = 100 \, \Omega $.
We need to find the resistance $ R_d $ for the dilute solution.
Steps to Solution:
The conductivity ratio for concentrated $(\kappa_c)$ and dilute $(\kappa_d)$ solutions relates to resistance:
$ \frac{\kappa_c}{\kappa_d} = \frac{R_d}{R_c} $
From the plot or given values (assuming values for molar conductivity):
$ \lambda_{m,c} $ and $ \lambda_{m,d} $ are the molar conductivities for the concentrated and dilute solutions.
If from the plot or data, $\lambda_{m,c} = 150 \, \text{S cm}^2 \text{mol}^{-1}$ and $\lambda_{m,d} = 100 \, \text{S cm}^2 \text{mol}^{-1}$.
Calculate the ratio of conductivities:
$ \frac{\kappa_c}{\kappa_d} = \frac{\left(\lambda_{m,c} \cdot C_c\right)}{\left(\lambda_{m,d} \cdot C_d\right)} $
Substituting values into the equation:
$ \frac{100 \cdot (0.15)^2}{150 \cdot (0.1)^2} = \frac{R_d}{100} $
Solving the equation gives:
$ R_d = 150 \, \Omega $
Thus, the resistance of the conductivity cell with the dilute solution is $ R_d = 150 \, \Omega $.
Given below are two statements :
1 M aqueous solutions of each of Cu(NO3)2, AgNO3, Hg2(NO3)2, Mg(NO3)2 are electrolysed using inert electrodes. Given: E0Ag+/Ag = 0.80 V, E0Hg22+/Hg = 0.79 V, E0Cu2+/Cu = 0.24 V and E0Mg2+/Mg = -2.37 V.
Statement (I) : With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu.
Statement (II) : Magnesium will not be deposited at the cathode instead oxygen gas will be evolved at the cathode.
In the light of the above statements, choose the most appropriate answer from the options given below :
Both Statement I and Statement II are incorrect
Statement I is incorrect but Statement II is correct
Statement I is correct but Statement II is incorrect
Both Statement I and Statement II are correct
On charging the lead storage battery, the oxidation state of lead changes from $x_1$ to $y_1$ at the anode and from $x_2$ to $y_2$ at the cathode. The values of $x_1, y_1, x_2, y_2$ are respectively :
The standard cell potential $\left(\mathrm{E}_{\text {cell }}^{\ominus}\right)$ of a fuel cell based on the oxidation of methanol in air that has been used to power television relay station is measured as 1.21 V . The standard half cell reduction potential for $\mathrm{O}_2\left(\mathrm{E}_{\mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\circ}\right)$ is 1.229 V .
Choose the correct statement :
$\mathrm{H}^{+}>\mathrm{Na}^{+}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}$
$\mathrm{H}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{K}^{+}>\mathrm{Na}^{+}$
$\mathrm{Mg}^{2+}>\mathrm{H}^{+}>\mathrm{Ca}^{2+}>\mathrm{K}^{+}>\mathrm{Na}^{+}$
$\mathrm{H}^{+}>\mathrm{Na}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{K}^{+}$
Match List - I with List - II :
| List - I (Applications) | List - II (Batteries/Cell) |
|---|---|
| (A) Transistors | (I) Anode - Zn/Hg; Cathode - HgO + C |
| (B) Hearing aids | (II) Hydrogen fuel cell |
| (C) Inverters | (III) Anode - Zn; Cathode - Carbon |
| (D) Apollo space ship | (IV) Anode - Pb; Cathode - Pb | PbO2 |
Choose the correct answer from the options given below :
(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
$\mathrm{O}_2$ gas will be evolved as a product of electrolysis of :
(A) an aqueous solution of $\mathrm{AgNO}_3$ using silver electrodes.
(B) an aqueous solution of $\mathrm{AgNO}_3$ using platinum electrodes.
(C) a dilute solution of $\mathrm{H}_2 \mathrm{SO}_4$ using platinum electrodes.
(D) a high concentration solution of $\mathrm{H}_2 \mathrm{SO}_4$ using platinum electrodes.
Choose the correct answer from the options given below :
For a Mg | Mg2+ (aq) || Ag+ (aq) | Ag the correct Nernst Equation is :
The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?
Molar conductivity increases sharply with increase in concentration.
Molar conductivity decreases sharply with increase in concentration.
A small increase in molar conductivity is observed at infinite dilution.
A small decrease in molar conductivity is observed at infinite dilution.
The standard reduction potential values of some of the p-block ions are given below. Predict the one with the strongest oxidising capacity.
$E^o_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +1.15 \text{ V}$
$E^o_{\text{Al}^{3+}/\text{Al}} = -1.66 \text{ V}$
$E^o_{\text{Pb}^{4+}/\text{Pb}^{2+}} = +1.67 \text{ V}$
$E^o_{\text{Tl}^{3+}/\text{Tl}} = +1.26 \text{ V}$
Based on the data given below :
$\begin{array}{ll} \mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}}^{\circ}=1.33 \mathrm{~V} & \mathrm{E}_{\mathrm{Cl}_2 / \mathrm{Cl}^{(-)}}^{\circ}=1.36 \mathrm{~V} \\ \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^0=1.51 \mathrm{~V} & \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V} \end{array}$
the strongest reducing agent is :
For the given cell
$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}_{(\mathrm{aq})}^{+} \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}_{(\mathrm{s})}$
The standard cell potential of the above reaction is Given:
$\begin{array}{lr} \mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^\theta=\mathrm{xV} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^\theta=\mathrm{yV} \\ \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^\theta=\mathrm{zV} \end{array}$
Standard electrode potentials for a few half cells are mentioned below :
$\begin{aligned} & \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=0.80 \mathrm{~V}, \mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=-2.37 \mathrm{~V} \end{aligned}$
Which one of the following cells gives the most negative value of $\Delta \mathrm{G}^{\circ}$ ?
$
\mathrm{FeO}_4^{2-} \xrightarrow{+2.0 \mathrm{~V}} \mathrm{Fe}^{3+} \xrightarrow{0.8 \mathrm{~V}} \mathrm{Fe}^{2+} \xrightarrow{-0.5 \mathrm{~V}} \mathrm{Fe}^0
$
In the above diagram, the standard electrode potentials are given in volts (over the arrow).
The value of $\mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{2+}}$ is :
Given below are two statements :
Statement (I) : Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure metal as a cathode.
Statement (II) : The rate of corrosion is more in alkaline medium than in acidic medium.
In the light of the above statements, choose the correct answer from the options given below :
Which of the following electrolyte can be used to obtain $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8$ by the process of electrolysis ?
A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A . The amount of the aluminium deposited at the cathode is __________ .
[Given : molar mass of aluminium and chlorine are $27 \mathrm{~g} \mathrm{~mol}^{-1}$ and $35.5 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively. Faraday constant $\left.=96500 \mathrm{C} \mathrm{~mol}^{-1}\right]$
The standard reduction potentials at $298 \mathrm{~K}$ for the following half cells are given below :
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}, \quad \mathrm{E}^{\circ}=1.33 \mathrm{~V}$
$\begin{array}{ll} \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^{\circ}=-0.04 \mathrm{~V} \\ \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni} & \mathrm{E}^{\circ}=-0.25 \mathrm{~V} \\ \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} & \mathrm{E}^{\circ}=1.40 \mathrm{~V} \end{array}$
Consider the given electrochemical reactions,
The number of metal(s) which will be oxidized be $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$, in aqueous solution is _________.
Explanation:
To determine the number of metals that will be oxidized by $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ in aqueous solution, we need to compare their standard reduction potentials with that of the given reduction potential of $\mathrm{Cr}_2\mathrm{O}_7^{2-}$.
The reduction reaction for $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ is:
$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6\mathrm{e}^- \rightarrow 2\mathrm{Cr}^{3+}+7\mathrm{H}_2\mathrm{O}, \quad \mathrm{E}^{\circ}=1.33 \mathrm{~V}$
This means that $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ has a strong tendency to get reduced (due to its high reduction potential of 1.33 V). Thus, any metal with a lower standard reduction potential than 1.33 V has the potential to be oxidized by $\mathrm{Cr}_2\mathrm{O}_7^{2-}$.
Given the standard reduction potentials:
$\begin{array}{ll} \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^{\circ}=-0.04 \mathrm{~V} \\ \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni} & \mathrm{E}^{\circ}=-0.25 \mathrm{~V} \\ \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} & \mathrm{E}^{\circ}=1.40 \mathrm{~V} \end{array}$
Now we compare these reduction potentials with that of $\mathrm{Cr}_2\mathrm{O}_7^{2-}$:
- For $\mathrm{Fe}^{3+}/\mathrm{Fe}$, $\mathrm{E}^{\circ}=-0.04 \mathrm{~V}$ (lower than 1.33 V)
- For $\mathrm{Ni}^{2+}/\mathrm{Ni}$, $\mathrm{E}^{\circ}=-0.25 \mathrm{~V}$ (lower than 1.33 V)
- For $\mathrm{Ag}^{+}/\mathrm{Ag}$, $\mathrm{E}^{\circ}=0.80 \mathrm{~V}$ (lower than 1.33 V)
- For $\mathrm{Au}^{3+}/\mathrm{Au}$, $\mathrm{E}^{\circ}=1.40 \mathrm{~V}$ (higher than 1.33 V)
From the comparison, we see that $\mathrm{Fe}$, $\mathrm{Ni}$, and $\mathrm{Ag}$ have standard reduction potentials lower than 1.33 V, meaning these metals can be oxidized by $\mathrm{Cr}_2\mathrm{O}_7^{2-}$, but $\mathrm{Au}$ cannot.
Therefore, the number of metals that will be oxidized by $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ in aqueous solution is: 3.
Explanation:
$\begin{aligned} & 1 \times 2=\frac{Q}{96500} \\\\ & Q=2 \times 96500 \mathrm{C} \\\\ & =1.93 \times 10^5 \mathrm{C}\end{aligned}$
$ \mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 $
The standard reduction potentials are given as below $\left(\mathrm{E}_{\text {red }}^0\right)$ :
$ \begin{aligned} & \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\circ}=+1.51 \mathrm{~V} \\\\ & \mathrm{E}_{\mathrm{CO}_2 / \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4}^{\circ}=-0.49 \mathrm{~V} \end{aligned} $
If the equilibrium constant of the above reaction is given as $\mathrm{K}_{\mathrm{eq}}=10^x$, then the value of $x=$ __________ (nearest integer)
Explanation:
$\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {op }}^{\circ}$ of anode $+\mathrm{E}_{\mathrm{RP}}^{\circ}$ of cathode
$ =0.49+1.51=2.00 \mathrm{~V} $
At equilibrium
$ \mathrm{E}_{\text {cell }}=0 \text {, } $
$ \mathrm{E}_{\text {cell }}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K} $
$2=\frac{0.0591}{10} \log K$
$\log K=338$
$ \begin{aligned} & 2 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2(\mathrm{~g}) \\\\ & {\left[\mathrm{H}^{+}\right]=1 \mathrm{M}, \mathrm{P}_{\mathrm{H}_2}=2 \mathrm{~atm}} \end{aligned} $
(Given : $2.303 \mathrm{RT} / \mathrm{F}=0.06 \mathrm{~V}, \log 2=0.3$ )
Explanation:
To find the potential of the given half cell at 298 K, we can use the Nernst equation, which in this scenario is simplified due to the standard conditions for the hydrogen ion concentration ($[\mathrm{H}^{+}] = 1 M$). The half-reaction involved is the reduction of hydrogen ions to hydrogen gas:
$2 \mathrm{H}_{(\mathrm{aq})}^+ + 2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2 (g)$
Given:
- Pressure of $\mathrm{H}_2 = 2 \text{ atm}$
- Hydrogen ion concentration $[\mathrm{H}^{+}] = 1 M$
- $2.303 \frac{\mathrm{RT}}{\mathrm{F}} = 0.06 V$ (a constant that simplifies calculation at 298 K)
- $\log 2 = 0.3$
The Nernst equation for this reaction under the given conditions becomes:
$\mathrm{E} = \mathrm{E}^{\circ} - \frac{0.06}{n} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{[\mathrm{H}^{+}]^2}$
Where $\mathrm{E}^{\circ} = 0.00 V$ is the standard electrode potential for the hydrogen half-cell, $n = 2$ is the number of electrons transferred, and the pressure of $\mathrm{H}_2$ is given as 2 atm. Inserting the values, we get:
$\mathrm{E}=0.00 - \frac{0.06}{2} \log \frac{2}{1^{2}} = -\frac{0.06}{2} \times 0.3 = -0.03 \times 0.3 = -0.009 = -0.9 \times 10^{-2} \text{V}$
Therefore, the potential of the given half-cell at 298 K is $-0.9 \times 10^{-2} \text{V}$.
$1.0 \times 10^{-16}, 1.2 \times 10,3.91,1.5 \times 10^{-2}, 1 \times 10^{-7}, 1.0 \times 10^3$.
The number of conductors among the materials is _____________.
Explanation:
Conductivity (S m$^{-1}$)
$\left.\begin{array}{l} 2.1 \times 10^3 \\ 1.2 \times 10 \\ 3.91 \\ 1 \times 10^3 \end{array}\right\} \text { conductors at } 298.15 \mathrm{~K}$
$1 \times 10^{-16} \text { Insulator at } 298.15 \mathrm{~K}$
$\left.\begin{array}{l} 1.5 \times 10^{-2} \\ 1 \times 10^{-7} \end{array}\right\} \text { Semiconductor at } 298.15 \mathrm{~K}$
Therefore number of conductors is 4.
Number of alkanes obtained on electrolysis of a mixture of $\mathrm{CH}_3 \mathrm{COONa}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{COONa}$ is ________.
Explanation:
$\mathrm{CH}_3 \mathrm{COONa} \rightarrow \dot{\mathrm{C}} \mathrm{H}_3$
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{COONa} \rightarrow \dot{\mathrm{C}}_2 \mathrm{H}_5$
$2 \dot{\mathrm{C}}_2 \mathrm{H}_5 \rightarrow \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$
$2 \dot{\mathrm{CH}}_3 \rightarrow \mathrm{CH}_3-\mathrm{CH}_3$
$\dot{\mathrm{C}} \mathrm{H}_3+\dot{\mathrm{C}}_2 \mathrm{H}_5 \rightarrow \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_3$
One Faraday of electricity liberates $x \times 10^{-1}$ gram atom of copper from copper sulphate. $x$ is ________.
Explanation:
To find the value of x when one Faraday of electricity liberates x times $10^{-1}$ gram atom of copper from copper sulphate, we need to understand how electricity interacts with copper ions in solution. The key reaction is:
$\mathrm{Cu}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}$
This shows that copper ions (Cu2+) gain two electrons (e-) to become copper metal (Cu). From electrochemistry, we know that:
- 2 Faradays of electricity are required to deposit 1 mole (or 1 gram atom) of copper.
- Therefore, 1 Faraday will deposit half of that amount, which is 0.5 mole, or in other terms, 0.5 gram atom of copper.
- Expressing 0.5 in the form of x times $10^{-1}$ gives us $5 \times 10^{-1}$.
This means x equals 5.