Electrochemistry

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: $\boxed{78\ \text{mL}}$

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} $

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).

$ \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} $

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} $

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} $

$ \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} $

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: $200\ \text{mV}$

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 $\Lambda_m^\infty = 6~\mathrm{S~m}^2 \mathrm{~mol}^{-1}$.

$\begin{aligned} & \mathrm{Cr}_2 \mathrm{O}_{7(\mathrm{aq})}^{-2}+14 \mathrm{H}_{(\mathrm{aq})}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{+3}+7 \mathrm{H}_2 \mathrm{O}_{(\ell)} \\ & \mathrm{E}_{\mathrm{R}}=\mathrm{E}_{\mathrm{R}}^0-\frac{0.059}{6} \log \frac{\left[\mathrm{Cr}^{+3}\right]^2}{\left[\mathrm{Cr}_2 \mathrm{O}_7^{-2}\right]\left[\mathrm{H}^{+}\right]^{14}} \\ & 0=1.33-\frac{0.059}{6} \log \frac{10^{-6}}{\left[\mathrm{H}^{+}\right]^{14}} \\ & \frac{1.33 \times 6}{0.059}=\log \frac{10^{-6}}{[\mathrm{H}]^{14}} \\ & 135.254=-6-14 \log \left[\mathrm{H}^{+}\right] \\ & 141.254=14 \mathrm{pH} \\ & \mathrm{pH}=\frac{141.254}{14}=10.08 \end{aligned}$

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 $=400 \mathrm{mV}$

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

$\boxed{23 \times 10^2\; \text{mS}\cdot \text{dm}^2/\text{mol}}.$

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.

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)

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 $.

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.

$\begin{aligned} & 2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \\\\ & \frac{\mathrm{W}}{\mathrm{E}}=\frac{\mathrm{Q}}{96500} \\\\ & \text { mole } \times \text { n-factor }=\frac{\mathrm{Q}}{96500}\end{aligned}$

$\begin{aligned} & 1 \times 2=\frac{Q}{96500} \\\\ & Q=2 \times 96500 \mathrm{C} \\\\ & =1.93 \times 10^5 \mathrm{C}\end{aligned}$

Cell $\mathrm{Rx}^{\mathrm{n}} ; \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2$

$\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {op }}^{\circ}$ 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} $

$2=\frac{0.0591}{10} \log K$

$\log K=338$

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 $-0.9 \times 10^{-2} \text{V}$.

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.

$\mathrm{CH}_3 \mathrm{COONa} \rightarrow \dot{\mathrm{C}} \mathrm{H}_3$

$\mathrm{C}_2 \mathrm{H}_5 \mathrm{COONa} \rightarrow \dot{\mathrm{C}}_2 \mathrm{H}_5$

$2 \dot{\mathrm{C}}_2 \mathrm{H}_5 \rightarrow \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$

$2 \dot{\mathrm{CH}}_3 \rightarrow \mathrm{CH}_3-\mathrm{CH}_3$

$\dot{\mathrm{C}} \mathrm{H}_3+\dot{\mathrm{C}}_2 \mathrm{H}_5 \rightarrow \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_3$

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.

$\begin{aligned} & \frac{\mathrm{W}}{\mathrm{E}}=\frac{\text { charge }}{1 \mathrm{~F}} \\ & \frac{1.314}{\frac{197}{3}}=\frac{\mathrm{Q}}{1 \mathrm{F}} \\ & \mathrm{Q}=2 \times 10^{-2} \mathrm{~F} \end{aligned}$

$\begin{aligned} & \mathrm{Zn}^{+2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} \\ & \mathrm{W}=\mathrm{Z} \times \mathrm{i} \times \mathrm{t} \\ & =\frac{65.4}{2 \times 96500} \times 0.015 \times 15 \times 60 \\ & =45.75 \times 10^{-4} \mathrm{gm}\end{aligned}$

$\begin{aligned} & 2 \mathrm{H}_{(\mathrm{aq} .)}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g}) \\ & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^0-\frac{0.059}{2} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2} \\ & =0-0.059 \times 3=-0.177 \text { volts. }=-17.7 \times 10^{-2} \mathrm{~V} \end{aligned}$

First, we need to determine the amount of $\mathrm{O}_2$ (oxygen gas) in moles that is displaced by the quantity of electricity mentioned. To do this, we'll use the molar volume of a gas at Standard Temperature and Pressure (S.T.P.), which is approximately $22.4 \mathrm{~L/mol}$ (or $22400 \mathrm{~mL/mol}$).

The volume of $\mathrm{O}_2$ is given as $5600 \mathrm{~mL}$. Now, we convert this volume to moles:

$ \text{moles of } \mathrm{O}_2 = \frac{\text{volume of } \mathrm{O}_2}{\text{molar volume}} = \frac{5600 \mathrm{~mL}}{22400 \mathrm{~mL/mol}} = 0.25 \mathrm{~mol} $

Next, we'll use Faraday's laws of electrolysis to relate the moles of $\mathrm{O}_2$ to the moles of silver being displaced. In electrolysis, silver is deposited at the cathode according to the following half-reaction:

$ \mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag} $

This tells us that for each mole of $\mathrm{Ag}^+$ ions, only 1 mole of electrons is required to reduce it to silver metal $\mathrm{Ag}$. However, the liberation of oxygen gas involves the following half-reaction:

$ 2\mathrm{H_2O} (l) \rightarrow \mathrm{O}_2 (g) + 4H^+ (aq) + 4e^- $

From this reaction, we can see that 1 mole of $\mathrm{O}_2$ gas requires 4 moles of electrons to be produced. Therefore, the number of moles of electrons associated with the $0.25 \mathrm{~mol}$ of $\mathrm{O}_2$ will be:

$ \text{moles of electrons for } \mathrm{O}_2 = 0.25 \mathrm{~mol} \times 4 = 1 \mathrm{~mol} $

Now, since it takes 1 mole of electrons to reduce 1 mole of $\mathrm{Ag}^+$. The moles of electrons required is equal to the moles of $\mathrm{Ag}$ produced. Hence, we have 1 mole of $\mathrm{Ag}$ being deposited.

Finally, to calculate the mass of this silver, we use the molar mass of silver:

$ \text{Mass of Ag} = (\text{moles of Ag}) \times (\text{Molar mass of Ag}) = 1 \mathrm{~mol} \times 108 \mathrm{~gmol}^{-1} = 108 \mathrm{~g} $

So, the mass of silver displaced by the quantity of electricity that displaces 5600 mL of $\mathrm{O}_2$ at S.T.P. will be $108 \mathrm{~g}$.

Let us evaluate each statement one by one to determine whether it is true or false:

(A) Conductivity always decreases with decrease in concentration for both strong and weak electrolytes.

This statement is true. Conductivity is defined as the ability of a solution to conduct electricity, and it depends on the concentration of ions in the solution. As the concentration of ions decreases, the conductivity of the solution also decreases, regardless of whether the electrolyte is strong or weak.

(B) The number of ions per unit volume that carry current in a solution increases on dilution.

This statement is false. The number of ions per unit volume in a solution decreases on dilution, because the total number of ions in the solution remains the same, but the volume increases.

(C) Molar conductivity increases with decrease in concentration.

This statement is generally true. Molar conductivity is a measure of the ability of an electrolyte to conduct electricity, and it depends on both the concentration of the electrolyte and the mobility of the ions. As the concentration of the electrolyte decreases, the molar conductivity increases because the ions become more separated from each other, and the electrostatic interactions between them become weaker, allowing them to move more freely in the solution.

(D) The variation in molar conductivity is different for strong and weak electrolytes.

This statement is true. The variation in molar conductivity with concentration is different for strong and weak electrolytes. Strong electrolytes dissociate completely in solution, so their molar conductivity increases rapidly with dilution, while weak electrolytes dissociate only partially, so their molar conductivity increases more slowly with dilution.

(E) For weak electrolytes, the change in molar conductivity with dilution is due to decrease in degree of dissociation.

This statement is false. For weak electrolytes, the change in molar conductivity with dilution is due to an increase in the degree of dissociation, not a decrease.

Therefore, there are $\boxed{3}$ correct statements in the given options, namely (A), (C), (D).

Given:

Standard reduction potential for Cu²⁺/Cu, E° = 0.34 V

Ksp of Cu(OH)₂ = 1 × 10⁻²⁰

2.303RT/F = 0.059 V

pH = 14

First, we have the solubility equilibrium for Cu(OH)₂:

$\mathrm{Cu}(\mathrm{OH})_2(\mathrm{~s}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})$

The Ksp expression for this reaction is:

$\mathrm{Ksp}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2$

At pH 14, the concentration of OH⁻ ions is 1 M:

$\left[\mathrm{OH}^{-}\right] = 1 \mathrm{M}$

Now we can find the concentration of Cu²⁺:

$\left[\mathrm{Cu}^{2+}\right]=\frac{\mathrm{Ksp}}{\left[\mathrm{OH}^{-}\right]^2}=\frac{1 \times 10^{-20}}{1^2}=10^{-20} \mathrm{M}$

The half-cell reaction for the reduction of Cu²⁺ is:

$\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})$

Now we can use the Nernst equation to calculate the reduction potential at pH 14:

$E = E° - \frac{0.059}{n} \log_{10} \frac{1}{\left[\mathrm{Cu}^{2+}\right]}$

Here, n = 2 (number of electrons transferred in the Cu²⁺/Cu couple).

$E = 0.34 - \frac{0.059}{2} \log_{10} \frac{1}{10^{-20}}$

$E = 0.34 - \frac{0.059}{2} \times 20$

$E = 0.34 - 0.59$

$E = -0.25 \mathrm{~V}$

Thus, the reduction potential at pH 14 for the Cu²⁺/Cu couple is -0.25 V. In terms of x × 10⁻² V:

$(-) x \times 10^{-2} \mathrm{~V} = -0.25 \mathrm{~V}$

The value of x is 25.

Using the Faraday's law of electrolysis, we can directly relate the amount of substance deposited (in this case, the nickel layer) with the electric charge passed through the electrolyte.

The formula for Faraday's law of electrolysis is:

$W = z \times i \times t$

where W is the amount of substance deposited (in grams), z is the electrochemical equivalent (grams per coulomb), i is the current (in amperes), and t is the time (in seconds).

By relating the density and volume of the nickel layer to the electric charge passed through the electrolyte, we can calculate the time needed for the deposition:

$10 \times 100 \times 0.0001 = \frac{\left(\frac{\text { atomic wt. }}{\text { v.f }}\right) \times 2 \times x}{96500}$

where v.f is the valence factor for the reaction (in this case, 2).

Solving for x, we get:

$x = 161 \, \mathrm{sec}$

So, the value of x is 161 seconds.

A. $\mathrm{E_{\text {cell }}}$ is an intensive parameter.

This statement is correct. The standard cell potential ($\mathrm{E}^{\ominus}_{\text{cell}}$) is an intensive parameter because it depends only on the nature of the reactants and the products, and not on the size or shape of the electrodes or the amount of material present.

B. A negative $\mathrm{E}^{\ominus}$ means that the redox couple is a stronger reducing agent than the $\mathrm{H}^{+} / \mathrm{H}_{2}$ couple.

This statement is also correct. A negative standard reduction potential ($\mathrm{E}^{\ominus}$) for a redox couple indicates that the couple is a stronger reducing agent than the standard hydrogen electrode ($\mathrm{H}^{+} / \mathrm{H}_{2}$), which has a standard reduction potential of 0 V.

C. The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction.

This statement is correct. The amount of electricity required for a particular oxidation or reduction reaction depends on the number of electrons involved in the reaction, which in turn depends on the stoichiometry of the electrode reaction.

D. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

This statement is also correct. Faraday's laws of electrolysis state that the amount of chemical reaction at an electrode during electrolysis is directly proportional to the amount of electricity passed through the electrolyte.

Therefore, all of the statements A, B, C, and D are correct.

In this problem, we're considering three different half-cell reactions involving lead ions.

1) The reduction of $\mathrm{Pb}^{2+}$ ions to lead metal :

$\mathrm{Pb}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}$

The Gibbs free energy change for this process can be written as $\Delta \mathrm{G}_1^0=-2 \mathrm{FE}_1^0$, where $\mathrm{E}_1^0$ is the standard cell potential, F is Faraday's constant, and the factor of 2 is because two electrons are involved in the reaction.

2) The reduction of $\mathrm{Pb}^{4+}$ ions to lead metal :

$\mathrm{Pb}^{4+} + 4 \mathrm{e}^{-} \rightarrow \mathrm{Pb}$

This reaction has a Gibbs free energy change of $\Delta \mathrm{G}_2^0=-4 \mathrm{FE}_2^0$.

3) The oxidation of $\mathrm{Pb}^{2+}$ ions to $\mathrm{Pb}^{4+}$ ions :

$\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{4+} + 2 \mathrm{e}^{-}$

The Gibbs free energy change for this process is $\Delta \mathrm{G}_3^0=-2 \mathrm{FE}_3^0$.

We can write the third reaction as the difference between the first two reactions (i.e., reaction 2 - reaction 1). This implies that the Gibbs free energy changes for these reactions should add up accordingly :

$\Delta \mathrm{G}_3^0=\Delta \mathrm{G}_2^0 - \Delta \mathrm{G}_1^0$

Substituting the expressions for the Gibbs free energy changes from the half-cell reactions into this equation, we get :

$-2 \mathrm{FE}_3^0 = -4 \mathrm{FE}_2^0 - (-2 \mathrm{FE}_1^0) = 2F (2n - m)$

Solving this equation for $E_3^0$ gives :

$E_3^0 = m - 2n$

However, the problem statement tells us that $E_3^0$ can also be written as $m - xn$. Comparing these two expressions for $E_3^0$, we see that $x$ must be equal to 2.

So, the value of $x$ is 2.

Given that the specific conductance, $k$, is $5 \times 10^{-5} ~S~cm^{-1}$ and the concentration, $C$, is $0.0025~M$, we can find the molar conductivity, $\lambda_m$, as follows:

$\lambda_m = \frac{k}{C} \times 1000 = \frac{5 \times 10^{-5} \times 10^3}{0.0025} = \frac{5 \times 10^{-2}}{2.5 \times 10^{-3}} = 20 ~S~cm^2~mol^{-1}$

Next, we find the degree of dissociation, $\alpha$, by dividing $\lambda_m$ by the limiting molar conductivity, $\lambda_m^o$:

$\alpha = \frac{20}{400} = \frac{1}{20}$

Finally, we use the formula for the dissociation constant of a weak acid, $K_a$:

$K_a = \frac{C \alpha^{2}}{1-\alpha} = \frac{0.0025 \times \frac{1}{20} \times \frac{1}{20}}{\frac{19}{20}} = \frac{0.0025}{19 \times 20} = 6.6 \times 10^{-6} = 66 \times 10^{-7}$

So, the correct answer is 66.

$ \begin{aligned} & \mathrm{FeO}_4^{2-} \stackrel{+2.2 \mathrm{~V}}{\longrightarrow} \mathrm{Fe}^{3+} ~~~\Delta \mathrm{G}_1=-6.6 \mathrm{~F} \\\\ & \mathrm{Fe}^{3+} \stackrel{+0.70 \mathrm{~V}}{\longrightarrow} \mathrm{Fe}^{2+} ~~~\Delta \mathrm{G}_2=-0.7 \mathrm{~F} \end{aligned} $

Hence for

$ \begin{aligned} & \mathrm{FeO}_4^{2-} \longrightarrow \mathrm{Fe}^{2+} \Delta \mathrm{G} =-7.3 \mathrm{~F} \\\\ & =-\mathrm{nEF} \\\\ & \mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{+2}}^0=\frac{-7.3 \mathrm{~F}}{-4 \mathrm{~F}}= 1.825, \mathrm{n}=4 \\\\ & =1825 \times 10^{-3} \mathrm{~V} \end{aligned} $

$n=$ electron exchange of that half cell reaction.

Let's go through the statements one by one:

A. The electrical work that a reaction can perform at constant pressure and temperature is equal to the reaction Gibbs energy.

This statement is correct. The maximum non-expansion work that a system can perform at constant temperature and pressure is given by the change in Gibbs free energy. This is especially relevant for electrochemical reactions where this non-expansion work appears as electrical work.

B. E°cell is dependent on the pressure.

This statement is incorrect. The standard cell potential, E°cell, is not dependent on pressure. It is dependent on the nature of the reactants and products (their identities and stoichiometric ratios), as well as temperature, but not pressure. The E°cell is calculated using standard conditions, which includes a standard pressure of 1 bar or approximately 1 atm.

C. $\frac{d E^{\theta} \text { cell }}{\mathrm{dT}}=\frac{\Delta_{\mathrm{r}} \mathrm{S}^{\theta}}{\mathrm{nF}}$

This statement is correct. This equation is derived from the Gibbs-Helmholtz equation, which describes the temperature dependence of the change in Gibbs free energy. In an electrochemical cell, the change in Gibbs free energy can be related to the cell potential.

D. A cell is operating reversibly if the cell potential is exactly balanced by an opposing source of potential difference.

This statement is correct. In a reversible electrochemical cell, the cell potential is exactly balanced by an external, opposing potential. This ensures that the reaction proceeds at an infinitesimally slow rate, allowing the system to maintain equilibrium at all times.

So, among these four statements, only one (Statement B) is incorrect.

To determine the metals that can be oxidized by $NO_3^-$, we need to find the metals that have a standard reduction potential (SRP) lower than 0.97 V (since $NO_3^-$ has an SRP of 0.97 V).

From the given data, we can find the metals that satisfy this condition:

1. $V^{2+} (\text{aq}) + 2e^- \rightarrow V$, $E^\theta = -1.19 \, \text{V}$ (less than 0.97 V, hence can be oxidized)


2. $Fe^{3+} (\text{aq}) + 3e^- \rightarrow Fe$, $E^\theta = -0.04 \, \text{V}$ (less than 0.97 V, hence can be oxidized)


3. $Ag^+ (\text{aq}) + e^- \rightarrow Ag (\text{s})$, $E^\theta = 0.80 \, \text{V}$ (less than 0.97 V, hence can be oxidized)


4. $Au^{3+} (\text{aq}) + 3e^- \rightarrow Au (\text{s})$, $E^\theta = 1.40 \, \text{V}$ (greater than 0.97 V, hence cannot be oxidized)

Therefore, $V$, $Fe$, and $Ag$ are the metals that can be oxidized by $NO_3^-$.

The total number of such metals is 3.

$ \begin{aligned} & \operatorname{AgBr}(\mathrm{S}) \rightleftharpoons \underset{\left(10^{-5}+\mathrm{x}\right)}{\rightleftharpoons \mathrm{Ag}^{+}}(\mathrm{aq})+\mathrm{Br}_{\mathrm{x}}^{-}(\mathrm{aq}) \\\\ & x\left(x+10^{-5}\right)=4.9 \times 10^{-13} \\\\ & x \simeq 4.9 \times 10^{-8} \mathrm{M} \\\\ & \lambda_{\mathrm{Ag}^{+}}^{\circ}=6 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\ & \lambda_{\mathrm{Br}^{-}}^0=8 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\ & \mathrm{~K}_{\text {solution }}=\mathrm{K}_{\mathrm{Ag}^{+}}+\mathrm{K}_{\mathrm{Br}^{-}}+\mathrm{K}_{\mathrm{NO}_3^{-}} \\\\ & =6 \times 10^{-3} \times 10^{-5} \times 10^3+8 \times 10^{-3} \times 4.9 \times 10^{-8} \times 10^3 \\\\ & +7 \times 10^{-3} \times 10^{-5} \times 10^3 \\\\ & =(6000+39.2+7000) \times 10^{-8} \\\\ & =13039.2 \times 10^{-8} ~\mathrm{Sm}^{-1} \end{aligned} $

$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightleftharpoons \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$

$ \begin{aligned} & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{5} \log \frac{\left[\mathrm{Mn}^{2+}\right]}{\left[\mathrm{MnO}_4^{-}\right]\left[\mathrm{H}^{+}\right]^8} \\\\ & \Rightarrow 1.282=1.54-\frac{0.059}{5} \log \frac{10^{-3}}{10^{-1} \times\left[\mathrm{H}^{+}\right]^8} \\\\ & \Rightarrow \frac{0.258 \times 5}{0.059}=\log \frac{10^{-2}}{\left[\mathrm{H}^{+}\right]^8} \\\\ & \Rightarrow 21.86=-2+8 \mathrm{pH} \\\\ & \therefore \mathrm{pH}=2.98 \\\\ & \simeq 3 \\\\ & \end{aligned} $

$ \begin{aligned} \text { Molar conductivity } & =\frac{\mathrm{k} \times 1000}{\mathrm{C}} \\\\ & =\frac{\frac{1}{5 \times 10^{-3}} \times 1000}{0.8} \\\\ & =\frac{10^6}{4}=0.25 \times 10^6 \\\\ & =25 \times 10^4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \end{aligned} $

Sol. $\Delta G^{\circ}=-R T \ell n K$

$ \begin{aligned} & -\mathrm{nFE}_{\text {cell }}^{\mathrm{o}}=-\mathrm{RT} \times 2.303\left(\log _{10} \mathrm{~K}\right) \\\\ & \frac{\mathrm{E}_{\text {Cell }}^{\mathrm{o}}}{0.06} \times \mathrm{n}=\log \mathrm{K} .......(1) \\\\ & \mathrm{Pd}^{+2} \text { (aq.) }+ \mathrm{2e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s}), \mathrm{E}_{\text {cat, }{ reduction}}^{\mathrm{o}}=0.83 \\\\ & \mathrm{Pd}(\mathrm{s})+4 \mathrm{Cl}^{-} \text {(aq.) } \rightleftharpoons \mathrm{PdCl}_4^{2-}, \text { (aq) }+2 \mathrm{e}^{-}, \mathrm{E}_{\text {Anode, Oxidation }}^0=0.65 \end{aligned} $

Net Reaction $\rightarrow \mathrm{Pd}^{2+}$ (aq.) $+4 \mathrm{Cl}^{-}$(aq.) $\rightleftharpoons \mathrm{PdCl}_4^{2-}$ (aq.)

$ \begin{aligned} & \mathrm{E}_{\text {cell }}^0=\mathrm{E}_{\text {cat,red }}^{\mathrm{o}}-\mathrm{E}_{\text {Anded, oxidid }}^0 \\\\ & \mathrm{E}_{\text {cell }}^{\mathrm{o}}=0.83-0.65 \\\\ & \mathrm{E}_{\text {cell }}^0=0.18 .........(2) \end{aligned} $

Also $\mathrm{n}=2$ .......(3)

Using equation (1), (2) and (3)

$\log K=6$

$ \begin{aligned} & \text { Cell reaction : } X+Y^{2+}(0.01 \mathrm{M}) \longrightarrow X^{2+}(0.001 \mathrm{M})+Y \\\\ & E_{\text {cell }}^o=0.36-(-2.36)=2.72 \mathrm{~V} \\\\ & E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{2.303 R T}{n F} \log \frac{\left[X^{2+}\right]}{\left[Y^{2+}\right]} \\\\ & =2.72-\frac{0.06}{2} \log \frac{(0.001)}{0.01} \quad[n=2 \text {, as two electrons are involved in the reaction] } \\\\ & =2.72-(0.03) \times(-1) \\\\ & =2.72+0.03 \\\\ & =2.75 \mathrm{~V}=275 \times 10^{-2} \mathrm{~V} \\\\ & \end{aligned} $

Anode ${H_2} \to 2{H^ + } + 2{e^ - }$

Cathode $(F{e^{3 + }} + {e^ - } \to F{e^{2 + }}) \times 2$

${H_2} + 2F{e^{3 + }} \to 2{H^ + } + 2F{e^{2 + }}$

${E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {\left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right)^2}$

$0.712 = 0.771 - 0.059\log {{F{e^{2 + }}} \over {F{e^{3 + }}}}$

$ - 0.059 = - 0.059\log {{F{e^{2 + }}} \over {F{e^{3 + }}}}$

${{[F{e^{2 + }}]} \over {[F{e^{3 + }}]}} = 10$

$E_{cell}^o = {{2.303\,RT} \over {2F}}\log k$

$E_{cell}^o = {{0.059} \over 2}\log ({10^{20}})$

$E_{Z{n^{2 + }}/Zn}^o + 0.76 = 0.59$

$E_{Z{n^{2 + }}/Zn}^o = 0.59 - 0.76$

$E_{Zn/Z{n^{2 + }}}^o = 0.17\,V$

(A) $\Lambda_{\mathrm{m}}^{\circ}$ for 'A' cannot be obtained by extra polation.

(C) At infinite dilution, value of degree of dissociation approaches one.

$\therefore \mathrm{A}$ and $\mathrm{C}$ are incorrect

Overall reaction :-

$ \begin{aligned} & \mathrm{H}_{2(\mathrm{~g})}+\mathrm{M}_{(\mathrm{aq})}^{3+} \longrightarrow \mathrm{M}_{\text {(aq) }}^{+}+2 \mathrm{H}_{(\mathrm{aq})}^{+} \\\\ & \mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cathode }}^{\mathrm{o}}-\mathrm{E}_{\text {anode }}^{\mathrm{o}}-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right] \times 1^2}{\left[\mathrm{M}^{+3}\right] 1} \\\\ & 0.1115=0.2-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\\\ & 3=\log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\\\ & \therefore \mathrm{a}=3 \end{aligned} $

$ \begin{aligned} & \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\mathrm{Fe}^{3+}(\mathrm{aq} .) \longrightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq} .) \\\\ & \mathrm{E}=\mathrm{E}^{\mathrm{o}}-\frac{0.059}{1} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} \\\\ & \Rightarrow \quad 0.712=(0.771-0)-\frac{0.059}{1} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} \\\\ & \Rightarrow \quad \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}=\frac{(0.771-0712)}{0.059}=1 \\\\ & \Rightarrow \quad \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}=10 \end{aligned} $

$ \begin{aligned} & \mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \\\\ & \mathrm{E}=1.33-\frac{0.059}{6} \log \frac{(0.1)^2}{\left(10^{-2}\right)\left(10^{-3}\right)^{14}} \\\\ & \mathrm{E}=1.33-\frac{0.059}{6} \times 42=0.917 \\\\ & \mathrm{E}=917 \times 10^{-3} \\\\ & \mathrm{x}=917 \end{aligned} $

Anode : $\mathrm{Cu}(\mathrm{s}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}$

Cathode : $\left.[\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(\mathrm{s})\right] 2$

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Cus(s) $+2 \mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$

$ \begin{aligned} & E_{\text {cell }}=E_{\text {cell }}^{0}-\frac{0.06}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}} \\ & 0.43=\mathrm{E}_{\text {cell }}^{0}-\frac{0.06}{2} \log \left(\frac{10^{-3}}{\left(10^{-2}\right)^{2}}\right) \\ & 0.43=E_{\text {cell }}^{0}-0.03 \log 10 \\ & \mathrm{E}_{\text {cell }}^{0}=0.46 \mathrm{~V} \\ & \mathrm{E}_{\text {cell }}^{0}=\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}-\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0} \\ & \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=(0.80-0.46)=0.34 \mathrm{~V}=34 \times 10^{-2} \end{aligned} $

$\frac{l}{A}=129 \mathrm{~m}^{-1}$

$\mathrm{KCl}$ solution $1 \Rightarrow 74.5 \,\mathrm{ppm}, \mathrm{R}_{1}=100 \Omega$

$\mathrm{KCl}$ solution $2 \Rightarrow 149 \,\mathrm{ppm}, \mathrm{R}_{2}=50 \Omega$

$ \begin{aligned} &\text { Here, } \frac{p p m_{1}}{p p m_{2}}=\frac{M_{1}}{M_{2}}=\left(\frac{w_{1 / M_{0}}}{V} \times \frac{V}{w_{2 / M_{0}}}\right) \\ &\frac{\Lambda_{1}}{\Lambda_{2}}=\frac{k_{1} \times \frac{1000}{M_{1}}}{k_{2} \times \frac{1000}{M_{2}}} \\ &=\frac{k_{1}}{k_{2}} \times \frac{M_{1}}{M_{2}} \\ &=\frac{50}{100} \times 2 \\ &=\frac{\Lambda_{1}}{\Lambda_{2}}=1000 \times 10^{-3} \\ &=1000 \end{aligned} $

For $\mathrm{Fe}_{3} \mathrm{O}_{4}$,

$ x=\frac{+8}{3} $

where x is oxidation state of Fe.

$ \mathrm{Fe}_{3} \mathrm{O}_{4}+8 \mathrm{H}^{+}+8 \mathrm{e}^{-} \longrightarrow 3 \mathrm{Fe}+4 \mathrm{H}_{2} \mathrm{O} $

Charge required $=\frac{8}{3} \times \mathrm{F}=\frac{8 \mathrm{~F}}{3} \simeq 3 \mathrm{~F}$

Among the pairs given, $\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}$ has negative reduction potential which is $-0.41 \mathrm{~V}$.

$\operatorname{Cr}$ (III) $\Rightarrow d^{3}$

Number of unpaired electrons $=3$

$\mu=\sqrt{3(3+2)}=\sqrt{15} \simeq 4$ B.M.