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.

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:

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:

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:

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:

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:

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 $108 \mathrm{~g}$.