Electrochemistry

Half-cell reactions are -

$A{g^ + }(aq) + e \to Ag(s)$

$Ag(s) + C{l^ - }(aq) \to AgCl(s) + e$

Cell reaction : $A{g^ + }(aq) + C{l^ - }(aq) \to AgCl(s)$

(1) The cell is : $Ag|AgCl(s)|C{l^ - }(aq)||A{g^ + }(aq)|Ag$

$A{g^ + }(aq) + C{l^ - }(aq) \to AgCl(s)$

$\therefore$ $\Delta {G^0} = \Delta G_r^0(AgCl) - \Delta G_r^0(A{g^ + }) - \Delta G_r^0(C{l^ - })$

or, $\Delta {G^0} = [ - 109 - 77 - ( - 129)]$ kJ mol^$-$1

or, $\Delta {G^0} = - 57$ kJ mol^$-$1

But, $\Delta {G^0} = - nF{E^0}$ or, $ - 57000 = - 1 \times 96500 \times {E^0}$

or, ${E^0} = {{57000} \over {96500}}$ or, ${E^0} = 0.59V$

The solubility equilibrium for AgCl is

$AgCl(s)$ $\rightleftharpoons$ $A{g^ + }(aq) + C{l^ - }(aq)$

For this reaction $E_{cell}^0 = - 0.59V$

$\therefore$ ${\log _{10}}{K_{sp}} = {{nF{E^0}} \over {RT}} = - {{0.59} \over {0.059}} = - 10$

(2) Amount of zinc added $ = {{6.539 \times {{10}^{ - 2}}} \over {65.39}} = {10^{ - 3}}$ mol

Therefore, the following reactions will occur :

$2A{g^ + }(aq) + 2e \to 2Ag(s)$ ;

$Zn(s) \to Z{n^{2 + }}(aq) + 2e$ ;

$2A{g^ + }(aq) + Zn(s) \to Z{n^{2 + }}(aq) + 2Ag(s)$ ;

By Nernst equation, ${E_{cell}} = E_{cell}^0 - {{0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$

At equilibrium, ${E_{cell}} = 0$, $\therefore$ $1.58 = {{0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$

or, $\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}} = {{1.58 \times 2} \over {0.059}} = 53.47$

$\therefore$ Equilibrium constant, $K = {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$

$\therefore$ $\log K = 53.47$ or, $K = {10^{53.47}}$

Very high value of K indicates that the reaction goes to almost completion. Solubility of $AgCl = \sqrt {{K_{sp}}} = \sqrt {{{10}^{ - 10}}} = {10^{ - 5}}$ M

$\therefore$ $[A{g^ + }] = {10^{ - 5}}M$. Hence, number of mole of Ag⁺ ions in 100 mL solution = 10^$-$6. Since, the reaction goes to almost completion, the amount of Ag formed = 10^$-$6 mol.

$\bullet$ The standard free energy for a given chemical reaction is equal to the difference between standard free energy of product and standard free energy of reactant.

$\bullet$ The solubility product of AgCl can be calculated using its equilibrium constant, K.

$K_{sp} =\frac{1}{K}$

(A) The chemical reaction is,

$A{g^ + }(aq.) + C{l^ - }(aq.) \to AgCl(s)$

$\mathrm{\Delta_r G^\circ(AgCl)\quad-109~kJ/mole}$

$\mathrm{\Delta_r G^\circ(Cl^-)\quad-129~kJ/mole}$

$\mathrm{\Delta_r G^\circ(Ag^+)\quad-77~kJ/mole}$

(i) To represent the above chemical reaction, we need to write the half-cell reactions first.

$Ag + {1 \over 2}C{l_2} \to AgCl$ .... (i)

$Ag \to A{g^ + } + {e^ - }$ ..... (ii)

${1 \over 2}C{l_2} + {e^ - } \to C{l^ - }$ ..... (iii)

$A{g^ + } + C{l^ - } \to AgCl$ ..... (i-ii-iii)

The cell can be represented as follows based on the above cell reaction.

$Ag|A{g^ + }||AgCl||C{l^ - }||C{l_2}|Pt$

(ii) To calculate E$^\circ$ for the cell, the formula that can be used is,

$\Delta G^\circ=-nFE^\circ$ ..... (i)

Here, $\Delta G^\circ$ denotes the standard free energy and n denotes the number of electrons transferred. The F is Faraday’s constant and E$^\circ$ represents the standard cell potential.

Given,

$\mathrm{\Delta_r G^\circ(AgCl)\quad-109~kJ/mole}$

$\mathrm{\Delta_r G^\circ(Cl^-)\quad-129~kJ/mole}$

$\mathrm{\Delta_r G^\circ(Ag^+)\quad-77~kJ/mole}$

For a given chemical reaction, the standard Gibbs free energy is the difference between the sum of Gibbs free energy of products and the sum of Gibbs free energy of reactants.

$\Delta G^\circ$ = [$\Delta G^\circ$ of AgCl - ($\Delta G^\circ$ of Ag$^+$ + $\Delta G^\circ$ of Cl$^-$)] ..... (ii)

Putting the respective values in equation (ii),

$\therefore$ $\Delta G^\circ=-109-(-129+77)=-57$ kJ/mole

For given chemical reaction, $n=1$, F = 96500C and $\Delta G^\circ=-57$ kJ/mole

Substituting the respective values in equation (i), we get

$-57=-1\times96500\times E^\circ$

$\Rightarrow E^\circ=\frac{57000}{96500}=0.59$ V

Therefore, the value of standard cell potential for the given reaction is 0.59 V

(iii) The K$_{sp}$ represents the solubility product constant for a substance, in this case AgCl.

To determine the K$_{sp}$, the formula that can be used is,

$K_{sp}=\frac{1}{K}$ ..... (iii)

Here, K is the equilibrium constant for the given chemical reaction.

The equilibrium constant can be determined by using the value of $\Delta G^\circ$.

$\Delta G^\circ = - 2.303\,RT\,\log K$ .... (iv)

$\Delta G^\circ = - {{57\,kJ} \over {mole}}$

$ = - 57000$ J/mole

R = 8.314 JK$^{-1}$ mol$^{-1}$

T = 298 K

Substitute the respective values in equation (iv), we get

$-57000=-2.303\times8.314\times298\times \log K$

$\therefore \log K=\frac{57000}{2.303\times8.314\times298}$

$=9.98 \approx 10$

$\therefore$ K = 10$^{10}$

Substitute the value of K in equation (iii) to calculate the value of K$_{sp}$ of AgCl.

${K_{sp}} = {1 \over {{{10}^{10}}}} = {10^{ - 10}}$

$\therefore$ ${\log _{10}}{K_{sp}} = - 10$

Therefore, the solubility product, K$_{sp}$ of AgCl is 10$^{-10}$.

(B) Given,

$\mathrm{Ag^++e^-\to Ag\quad E^\circ=0.80~V}$

$\mathrm{Zn^{2+}+2e^-\to Zn\quad E^\circ=-0.76~V}$

Mass of Zn = 6.539 $\times$ 10$^{-2}$ g

Atomic mass unit of Zn = 65.39

Volume of AgCl = 100 ml

To find the moles of Zn added, divide the mass of metallic Zn by atomic mass unit of Zn.

Moles of Zinc added $=\frac{6.539\times10^{-2}}{65.39}=10^{-3}$ moles

From the given half reaction, overall reaction can be represented and value of E$^\circ$ can be calculated.

The moles of metallic Zn added is 10–3 moles and from the given reaction, the number of moles of Ag added will be twice the number of moles of Zn added.

Therefore, the number of moles of Ag$^+$ added is 10$^{–6}$ moles.

The value of ${{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$ can be calculated using Nernst equation.

${E_{cell}} = E_{cell}^o - {{0.0591} \over n}{\log _{10}}{{[Z{n^{ + 2}}]} \over {{{[A{g^ + }]}^2}}}$ ..... (v)

At equilibrium, ${E_{cell}} = 0$

For the given reaction, $n = 2$ and $E^\circ = 1.56$ V

Substituting the respective values in equation (v),

$0 = 1.56 - {{0.0591} \over 2}{\log _{10}}{{[Z{n^{ + 2}}]} \over {{{[A{g^ + }]}^2}}}$

${\log _{10}}{{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}} = {{1.56 \times 2} \over {0.0591}}$

$\therefore$ ${\log _{10}}{{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}} = 52.8$

Hence, the value of ${{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$ is 52.8.

The equilibrium constant K can be calculated as follows :

As $K = {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$

$\therefore$ $K = {10^{52.8}}$

Since the value of equilibrium constant is very high, the reaction almost goes to completion. Therefore, almost 100% Ag precipitate out. Thus, moles of Ag formed during the reaction will be 10$^{–5}$ moles.

[$\because$ ${K_{sp}}[AgCl] = {10^{ - 10}}$]

Final Answer :

(A) (i) $Ag|A{g^ + }||AgCl||C{l^ - }||C{l_2}|Pt$

(ii) ${E^0} = 0.59$ V

(iii) ${\log _{10}}{K_{sp}} = -10$

(B) ${\log _{10}} = {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}} = 52.8$

Moles of Ag formed $ = {10^{ - 6}}$ moles.

We know, $\Delta {G^0} = - nF{E^0}$

(1) $C{u^{2 + }} + e \to C{u^ + }$ ; $\Delta G_1^0 = - 0.15F$

(2) $I{n^{2 + }} + e \to I{n^ + }$ ; $\Delta G_2^0 = + 0.4F$

(3) $I{n^{3 + }} + 2e \to I{n^ + }$ ; $\Delta G_3^0 = + 0.84F$

Adding equation (1) and (2) and subtracting equation (3) we get, $C{u^{2 + }} + I{n^{2 + }} \to C{u^ + } + I{n^{3 + }}$

$\Delta G_{}^0 = \Delta G_1^0 + \Delta G_2^0 - \Delta G_3^0 = - 0.59F$

$\Delta G_{}^0 = - 2.303RT\log {K_{eq}}$ $\therefore$ $ - 0.59F = - 2.303RT\log {K_{eq}}$

or, $\log {K_{eq}} = {{0.59 \times 96500} \over {2.303 \times 8.314 \times 298}} \approx 10$ $\therefore$ ${K_{eq}} = {10^{10}}$

The two given cells are represented as :

$Zn(s)|Z{n^{2 + }}({C_1})||C{u^{2 + }}(aq)(C = ?)|Cu(s)$, ${E_{cell}} = {E_1}$

$Zn(s)|Z{n^{2 + }}({C_2})||C{u^{2 + }}(aq)(C = 0.5\,M)|Cu(s)$, ${E_{cell}} = {E_2}$

Given, E₂ > E₁ , E₂ $-$ E₁ = 0.03 and C₁ = C₂ [concentration of Zn²⁺ is the same in both the solutions]

$\therefore$ Cell reaction : $Zn(s) + C{u^{2 + }}(aq)$ $\rightleftharpoons$ $Z{n^{2 + }}(aq) + Cu(s)$

$\therefore$ ${E_{cell}} = E_{cell}^0 - {{2.303RT} \over {2F}}\log {{[Z{n^{2 + }}]} \over {[C{u^{2 + }}]}}$

For cell 1, ${E_1} = E_{cell}^0 - {{0.06} \over 2}\log {{{C_1}} \over C}$ [Given : ${{2.303RT} \over F} = 0.06$]

For cell 2, ${E_2} = E_{cell}^0 - {{2.303RT} \over {nF}}\log {{{C_2}} \over {0.05}}$

or, ${E_2} = E_{cell}^0 - {{0.06} \over 2}\log {{{C_2}} \over {0.05}}$

$\therefore$ ${E_2} - {E_1} = \left( {E_{cell}^0 - {{0.06} \over 2}\log {{{C_2}} \over {0.05}}} \right) - \left( {E_{cell}^0 - {{0.06} \over 2}\log {{{C_1}} \over C}} \right)$

or, $0.03 = {{0.06} \over 2}\left( {\log {{{C_2}} \over C} \times {{0.5} \over {{C_1}}}} \right) = {{0.06} \over 2}\log {{0.5} \over C}$ [$\because$ C₁ = C₂]

or, $\log {{0.5} \over C} = {{0.03 \times 2} \over {0.06}}$ or, ${{0.5} \over C} = 10$

$\therefore$ C = 0.05 M

Given cell : $Pt|{H_2}(g)|HCl(aq)|AgCl(s)|Ag(s)$

(1) The half-cell reactions are as follows:

At anode : ${1 \over 2}{H_2}(g) \to {H^ + }(aq) + e$

At cathode : $AgCl(s) + e \to Ag(s) + C{l^ - }(aq)$

Cell reaction : ${1 \over 2}{H_2}(g) + AgCl(s) \to {H^ + }(aq) + Ag(s) + C{l^ - }(aq)$

(2) $\Delta {S^0} = nF\left( {{{d{E^0}} \over {dT}}} \right)$, where n = Number of electrons involved in the cell reaction, F = Faraday = 96500 C, dE⁰ = Difference of standard electrode potential at two different temperatures = (0.21 $-$ 0.23) = $-$ 0.02 V and dT = difference of two temperatures = (308 $-$ 288) K = 20 K

$\therefore$ $\Delta {S^0} = 1 \times 96500 \times \left( {{{ - 0.02} \over {20}}} \right) = - 96.5$ J K^$-$1 mol^$-$1

We know, $\Delta {G^0} = - nF{E^0}$

$\therefore$ $\Delta G_{15^\circ \,C}^0 = - 1 \times 96500 \times 0.23$ [$\because$ $\Delta E_{15^\circ \,C}^0 = 0.23\,V$]

= $-$ 22195 J . mol^$-$1

$\therefore$ $\Delta H_{}^0 = \Delta G_{}^0 - T\Delta S_{}^0$

$ = - 22195 - 288 \times ( - 96.5) = - 49987$ J mol^$-$1

(3) Given $\Delta E_{(15^\circ \,C)}^0 = 0.23\,V$ and $\Delta E_{(35^\circ \,C)}^0 = 0.21\,V$

$\therefore$ ${{\Delta {E^0}} \over {\Delta T}} = {{(0.21 - 0.23)} \over {20}} = - 0.01$

$\therefore$ $\Delta$E⁰ for 10$^\circ$C = $-$0.01 $\times$ 10 = $-$0.1

$\therefore$ $\Delta E_{(25^\circ \,C)}^0 = \Delta E_{(15^\circ \,C)}^0 + ( - 0.1) = 0.23 - 0.1 = 0.22\,V$

$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0$

$0.22\,V = E_{C{l^ - }|AgCl|Ag}^0 - E_{2{H^ + }|{H_2}}^0 = E_{C{l^ - }|AgCl|Ag}^0 - 0$

$\therefore$ $E_{C{l^ - }|AgCl|Ag}^0$ = 0.22 V and $E_{A{g^ + }|Ag}^0$ = 0.80 V (given)

$E_{C{l^ - }|AgCl|Ag}^0 = E_{A{g^ + }|Ag}^0 - {{0.059} \over 1} \times \log {K_{sp}}(AgCl)$

or, $0.22\,V = 0.80\,V - {{0.059} \over 1}\log {K_{sp}}(AgCl)$

$\therefore$ ${K_{sp}}(AgCl) = 1.47 \times {10^{ - 10}}$

and ${K_{sp}}(AgCl) = [A{g^ + }] \times [C{l^ - }]$

$\therefore$ ${[A{g^ + }]^2} = 1.47 \times {10^{ - 10}}$

or, $[A{g^ + }] = \sqrt {1.47 \times {{10}^{ - 10}}} = 1.21 \times {10^{ - 5}}$

Quantity of charge passed = 2 $\times$ 10^$-$3 $\times$ 16 $\times$ 60 = 1.92 C

1.92 C charge $ \equiv {1 \over {96500}} \times 1.92 \equiv 1.99 \times {10^{ - 5}}$ mol of electrons

In the reaction, $C{u^{2 + }}(aq) + 2e \to Cu(s)$

2 mol of electrons = 1 mol of Cu²⁺ ions discharged

$\therefore$ 1.99 $\times$ 10^$-$5 mol of electrons = 9.95 $\times$ 10^$-$6 of Cu²⁺ ions discharged.

Absorbance of a solution is directly proportional to the concentration of the solution. 50% decrease of absorbance of the solution means 50% of the concentration of the solution is reduced. Hence, initial number of mol of Cu²⁺ ions = 2 $\times$ 9.95 $\times$ 10^$-$6 = 1.99 $\times$ 10^$-$5 mol.

$\therefore$ Initial concentration of Cu²⁺

i.e., $CuS{O_4} = {{1.99 \times {{10}^{ - 5}}} \over {250}} \times 1000\,M = 7.96 \times {10^{ - 5}}\,M$

For electrochemical cell,

$Pt(1)|F{e^{3 + }}$ , $Fe(a = 1)||C{e^{4 + }}$ , $C{e^{3 + }}(a = 1)|Pt(2)$

$E_{F{e^{3 + }}|F{e^{2 + }}}^0 = 0.77\,V$, $E_{C{e^{4 + }}|C{e^{3 + }}}^0 = 1.61\,V$

$\therefore$ $E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = 1.61 - 0.77 = 0.84\,V$

Since, $E_{cell}^0$ is +ve, the cell reaction will occur spontaneously.

Hence, the flow of current will be from cathode to anode (right to left), and the current will decrease with time.

Due to the passage of current, the cell reaction is :

$Ag(s) + {1 \over 2}C{u^{2 + }}(aq) \to A{g^ + }(aq) + {1 \over 2}Cu(s)$ ...... [1]

From Faraday's first law, $W = Z \times I \times t$

or, $W = {{equivalent\,weight\,(E)} \over {96500}} \times I \times t$

$\therefore$ ${W \over E}$ (gram equivalent) of $C{u^{2 + }} = {1 \over {96500}} \times I \times t$

$ = {1 \over {96500}} \times 9.65 \times 1 \times 60 \times 60 = 0.36$

$\therefore$ Decrease in concentration of $C{u^{2 + }} = {{0.36} \over 2}(M) = 0.18\,(M)$

$\therefore$ Remaining concentration of copper = 1 $-$ 0.18 = 0.82 (M)

$\therefore$ Increase in the concentration of silver ion = 0.36 (M)

$\therefore$ Concentration of silver ion = (1 + 0.36) = 1.36 (M)

For the reaction (1), ${E_{cell}} = E_{cell}^0 - {{0.059} \over 1}\log {{[A{g^ + }]} \over {{{[C{u^{2 + }}]}^{1/2}}}}$

Before passing current, ${E_{cell}} = E_{cell}^0 - {{0.059} \over 4}\log {1 \over {{1^2}}} = E_{cell}^0\,V$

When passage of current is stopped,

$E{'_{cell}} = E_{cell}^0 - {{0.059} \over n}\log {{[A{g^ + }]} \over {{{[C{u^{2 + }}]}^{1/2}}}}$

or, $E{'_{cell}} = E_{cell}^0 - {{0.059} \over 1}\log {{1.36} \over {{{(0.81)}^{1/2}}}} = (E_{cell}^0 - 0.010)\,V$

$\therefore$ $\Delta$E = change in cell potential $ = E{'_{cell}} - {E_{cell}} = - 0.010\,V$.

Given cell : $Ag|A{g^ + }(sat\,A{g_2}Cr{O_4}\,sol)||Ag(0.1\,M)|Ag$ and ${E_{cell}} = 0.164\,V$ at 298 K.

At anode : $Ag(s) \to A{g^ + }(sat\,A{g_2}Cr{O_4}) + e$

At cathode : $A{g^ + }(aq) + e \to Ag(s)$

Cell reaction : $A{g^ + }(aq) \to A{g^ + }(sat\,A{g_2}Cr{O_4})$

Let molar concentration of Ag⁺ ions in sat Ag₂CrO₄ be C₁M

$\therefore$ ${E_{cell}} = E_{cell}^0 - {{0.059} \over 1}\log {{{C_1}} \over {[A{g^ + }]}}$

$E_{cell}^0 = 0$ as both the electrodes are the same

$\therefore$ $0.164 = {{0.059} \over 1}\log {{0.1} \over {{C_1}}}$ or, $\log {{0.1} \over {{C_1}}} = {{0.164} \over {0.059}} = 2.774$

or, ${C_1} = {[A{g^ + }]_{A{g_2}Cr{O_4}}} = 1.66 \times {10^{ - 4}}(M)$

In, $A{g_2}Cr{O_4}(s)$ $\rightleftharpoons$ $2A{g^ + }(aq) + CrO_4^{2 - }(aq)$

$[CrO_4^{2 - }] = {{[A{g^ + }]} \over 2} = {{1.66 \times {{10}^{ - 4}}} \over 2}(M) = 0.83 \times {10^{ - 4}}(M)$

$\therefore$ K_sp of $A{g_2}Cr{O_4} = {[A{g^ + }]^2}[CrO_4^{2 - }]$

$ = (1.66 \times {10^{ - 4}}) \times 0.83 \times {10^{ - 4}} = 2.287 \times {10^{ - 12}}$

Given, $E_{F{e^{3 + }}|F{e^{2 + }}}^0 = 0.78V$ and $E_{I_3^ - |{I^ - }}^0 = 0.54V$

The half-cell reactions are as follows:

At anode : $3{I^ - } \to I_3^ - + 2e$, ${E^0} = 0.54$

At cathode : $2F{e^{3 + }} + 2e \to 2F{e^{2 + }}$, ${E^0} = + 0.78V$

Cell reaction : $2F{e^{3 + }} + 3{I^ - }$ $\rightleftharpoons$ $2F{e^{2 + }} + I_3^ - $

$\therefore$ $E_{cell}^0 = (0.78 - 0.54) = 0.24V$

We know, $E_{cell}^0 = {{0.059} \over 2}\log {K_{eq}}$ or, $0.24 = 0.029\log {K_{eq}}$

or, $\log {K_{eq}} = {{0.24} \over {0.029}} = 8.27 \simeq 8.00$ $\therefore$ ${K_{eq}} = {10^8}$

The half-cell reactions are as follows:

At anode : $F{e^{2 + }}(aq) \to F{e^{3 + }}(aq) + e$ ; ${E^0} = - 0.68\,V$

At cathode : $C{e^{4 + }}(aq) + e \to C{e^{3 + }}(aq)$ ; ${E^0} = 1.44\,V$

Cell reaction : $F{e^{2 + }}(aq) + C{e^{4 + }}(aq)$ $\rightleftharpoons$ $F{e^{3 + }}(aq) + C{e^{3 + }}(aq)$ ; $E_{cell}^0 = + 0.76\,V$

We know, $E_{cell}^0 = {{0.059} \over n}\log {K_{eq}}$ or, $ + 0.76 = {{0.059} \over 1}\log {K_{eq}}$

or, $\log {K_{eq}} = {{0.76} \over {0.059}} = 12.859$ $\therefore$ ${K_{eq}} = 7.227 \times {10^{12}}$

Charges passed through the cell = 8.46 $\times$ 3 $\times$ 3600 = 243648 coulombs

Electrode reaction : $A{g^ + }(aq) + e \to Ag(s)$ ..... [1]

243648 coulombs $ \equiv {{243648} \over {96500}} \equiv 2.524$ mol of electrons

According to equation (1) 2.524 mol of electrons = 180 $\times$ 2.524 = 272.6 g of Ag

$\therefore$ Volume of Ag plated out $ = {W \over {density}} = {{272.68} \over {10.5}} = 25.969$ cm³

Given, thickness of the Ag plating = 0.0254 cm

$\therefore$ Area of tray plated $ = {{25.969} \over {0.0254}} = 1022.4 = 1.02 \times {10^3}$ cm²

Given, $E_{C{u^{2 + }}|Cu}^0(aq) = + 0.34V$

$Cu{(OH)_2}(s) \to C{u^{2 + }}(aq) + 2O{H^ - }(aq)$, ${K_{sp}} = [C{u^{2 + }}]{[O{H^ - }]^2}$

Given, pH = 14, [H⁺] = 10^$-$14 M, [OH^$-$] = 1 M

K_sp of $Cu{(OH)_2} = 1.0 \times {10^{ - 19}}$

$\therefore$ $1.0 \times {10^{ - 19}} = [C{u^{2 + }}]{[O{H^ - }]^2}$

or, $[C{u^{2 + }}]{[1]^2} = 1.0 \times {10^{ - 19}}$ [$\because$ $C{u^{2 + }} + 2e \to Cu$]

$\therefore$ $[C{u^{2 + }}] = {10^{ - 19}}M$

$\therefore$ ${E_{C{u^{2 + }}|Cu}} = E_{C{u^{2 + }}|Cu}^0 - {{0.059} \over 2}\log {1 \over {[C{u^{2 + }}]}}$

$\therefore$ ${E_{C{u^{2 + }}|Cu}} = 0.34 - {{0.059} \over 2}\log {1 \over {({{10}^{ - 19}})}} = - 0.22V$

Given : $2Hg(l) + 2F{e^{3 + }}(aq) \to Hg_2^{2 + }(aq) + 2F{e^{2 + }}(aq)$

$[F{e^{3 + }}] = 1.0 \times {10^{ - 3}}(M)$

$\therefore$ [Fe³⁺] at equilibrium = 5% of 1.0 $\times$ 10^$-$3 (M)

$ = {5 \over {100}} \times 1.0 \times {10^{ - 3}}(M) = 5 \times {10^{ - 5}}(M)$

$\therefore$ $[F{e^{2 + }}] = (1.0 \times {10^{ - 3}}) - (5 \times {10^{ - 5}}) = 0.95 \times {10^{ - 3}}$

$\therefore$ $[Hg_2^{2 + }]$ at equilibrium $ = {{0.95 \times {{10}^{ - 3}}} \over 2}M$

$\therefore$ ${E_{cell}} = E_{cell}^0 - {{0.059} \over 2}\log {{[Hg_2^{2 + }]{{[F{e^{2 + }}]}^2}} \over {{{[F{e^{3 + }}]}^2}}}$

At equilibrium, ${E_{cell}} = 0$

$\therefore$ $E_{cell}^0 = {{0.059} \over 2}\log {{\left[ {{{0.95 \times {{10}^{ - 3}}} \over 2}} \right]{{[0.95 \times {{10}^{ - 3}}]}^2}} \over {{{[5 \times {{10}^{ - 5}}]}^2}}}$

or, $E_{cell}^0 = {{0.0591} \over 2}\log {{{{[95]}^3} \times {{10}^{ - 5}}} \over {50}} = - 0.0276$

Also, $E_{cell}^0 = E_{F{e^{3 + }}|F{e^{2 + }}}^0 - E_{Hg_2^{2 + }|Hg}^0$

or, $ - 0.0276 = 0.77 - E_{Hg_2^{2 + }|Hg}^0$

or, $E_{Hg_2^{2 + }|Hg}^0 = 0.77 + 0.0276 = 0.7976\,V$

Given, $E_{FeO|Fe}^0 = - 0.87\,V$, $E_{N{i_2}{O_3}|NiO}^0 = + 0.40\,V$

At anode : $Fe(s) + 2O{H^ - }(l) \to FeO(s) + {H_2}O(l) + 2e$

At cathode : $N{i_2}{O_3}(s) + {H_2}O(l) + 2e \to 2NiO(s) + 2O{H^ - }(aq)$

(1) Cell reaction : $Fe(s) + N{i_2}{O_3}(s) \to FeO(s) + 2NiO(s)$

(2) $E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = 0.40 - ( - 0.87) = + 1.27\,V$

${E_{cell}} = E_{cell}^0 - {{0.059} \over 2}\log {{[FeO]{{[NiO]}^2}} \over {[Fe][N{i_2}{O_3}]}}$

$ = {E^0} - {{0.059} \over 2}\log {{1 \times {1^2}} \over {1 \times 1}} = {E^0}$ [$\because$ Molar concentration of pure solid = 1]

The expression for E_cell does not contain concentration term of OH^$-$, so E_cell is independent of OH^$-$ concentration.

(3) Electrical energy obtained from 1 mole of

$N{i_2}{O_3} = n \times E_{cell}^0 \times F = 2 \times 1.27 \times 96500\,J$ = 245110 J = 245.11 kJ.

Given, $E_{A{g^ + }|Ag}^0 = 0.799$ (at 298 K) and K_sp of $AgI = 8.7 \times {10^{ - 17}}$

In a saturated solution of AgI, [Ag⁺] = [I^$-$]

$\therefore$ $[A{g^ + }] = \sqrt {{K_{sp}}} = \sqrt {8.7 \times {{10}^{ - 17}}} = 9.33 \times {10^{ - 9}}$

According to Nernst equation of $A{g^ + } + e \to Ag(s)$

$\therefore$ $E_{A{g^ + }|Ag}^{} = E_{A{g^ + }|Ag}^0 - {{0.059} \over n}\log {1 \over {[A{g^ + }]}}$

or, $E_{A{g^ + }|Ag}^{} = 0.799 - {{0.059} \over 1}\log {1 \over {[A{g^ + }]}}$

or, $E_{A{g^ + }|Ag}^{} = 0.799 - 0.059\log {1 \over {9.33 \times {{10}^{ - 9}}}}$

$\therefore$ $E_{A{g^ + }|Ag}^{} = 0.325$ V

The relation between $E_{{I^ - }|AgI|Ag}^0$ and $E_{A{g^ + }|Ag}^0$ is

$E_{{I^ - }|AgI|Ag}^0 = E_{A{g^ + }|Ag}^0 + 0.059\log {K_{sp}}$

$\therefore$ $E_{{I^ - }|AgI|Ag}^0 = 0.799 + 0.059\log (8.7 \times {10^{ - 17}}) = - 0.148$ V

Reaction :

$NO_3^ - (aq) + 2{H^ + }(aq) + e \to N{O_2}(g) + {H_2}O(l)$

(1) ${E_{cell}} = E_{cell}^0 - {{0.059} \over n}\log {{[N{O_2}][{H_2}O]} \over {[NO_3^ - ]{{[{H^ + }]}^2}}}$

$ = 0.78 - {{0.059} \over 1}\log {{1 \times 1} \over {1 \times {{(8)}^2}}} = 0.887$ V

(2) In neutral solution, [H⁺] = 10^$-$7 M

$\therefore$ $E = 0.78 - {{0.059} \over 1}\log {{1 \times 1} \over {1 \times {{({{10}^{ - 7}})}^2}}} = - 0.046$ V

$Cr{O_3}(aq) + 6{H^ + }(aq) + 6e \to Cr(s) + 3{H_2}O$

52g of Cr is deposited by 6 $\times$ 96500 C of electric current. [$\because$ atomic mass of Cr = 52 g mol^$-$1]

(1) Given, 24000 C current is passed.

$\therefore$ Amount of Cr plated out = ${{52 \times 24000} \over {6 \times 96500}}$ = 2.155 g

(2) According to Faraday's 1st law, W = Z $\times$ I $\times$ t

$\therefore$ 1.5 = Z $\times$ 12.5 $\times$ t

or, $1.5 = {{52} \over {6 \times 96500}} \times 12.5 \times t$

or, $t = {{1.5 \times 6 \times 96500} \over {52 \times 12.5}} = 1336.15$

(1) $2C{l^ - }(aq) + 2{H_2}O \to 2O{H^ - }(aq) + {H_2}(g) + C{l_2}(g)$

Cathode reaction : $2{H_2}O + 2e \to 2O{H^ - } + {H_2}$

Anode reaction : $2C{l^ - } \to C{l_2} + 2e$

(2) According to Faraday's first law, $W = Z \times It$ ...... (1)

Current efficiency = 62% = 0.62

Effective electric current (I) = 25 $\times$ 0.62 A

Mass of $C{l_2} = {10^3}g$ $\therefore$ Z for $C{l_2} = {{eq.\,weight} \over {96500}} = {{35.5} \over {96500}}$

Substituting in equation (1), ${10^3} = {{35.5} \over {96500}} \times 25 \times 0.62 \times t$

or, $t = {{{{10}^3} \times 96500} \over {35.5 \times 25 \times 0.62}} = 175374.83$ sec = 48.71 hrs

(3) No. of equivalents of OH^$-$ = no. of equivalents of Cl₂ = ${{{{10}^3}} \over {35.5}} = 28.17$ (equivalent weight of Cl₂ = 35.5)

Equivalent weight and formula mass of OH^$-$ are equal.

$\therefore$ Number of moles of OH^$-$ = 28.17

$\therefore$ $[O{H^ - }] = {{number\,of\,moles} \over {volume\,(in\,L)}} = {{28.17} \over {20}} = 1.408\,M$

For the galvanic cell,

$Ag|AgCl(s)$, $KCl(0.2M)||KBr(0.001\,M)$, $AgBr(s)|Ag$

${E_{cell}} = {{0.591} \over n}\log {{[A{g^ + }]AgBr} \over {[A{g^ + }]AgCl}}$ ...... (1)

Given, ${K_{sp}}(AgCl) = 2.8 \times {10^{ - 10}}$. Now, [Cl^$-$] = 0.2 (M)

$\therefore$ $2.8 \times {10^{ - 10}} = {[A{g^ + }]_{AgCl}} \times 0.2$

or, ${[A{g^ + }]_{AgCl}} = {{2.8 \times {{10}^{ - 10}}} \over {0.2}} = 14 \times {10^{ - 10}}(M)$

Given, ${K_{sp}}(AgBr) = 3.3 \times {10^{ - 13}}$. Here, [Br^$-$] = 0.001 M

$\therefore$ $3.3 \times {10^{ - 13}} = {[A{g^ + }]_{AgBr}} \times 0.001$

or, ${[A{g^ + }]_{AgBr}} = {{3.3 \times {{10}^{ - 13}}} \over {0.001}} = 3.3 \times {10^{ - 10}}$ (M)

Substituting in equation (1), ${E_{cell}} = {{0.059} \over 1}\log {{3.3 \times {{10}^{ - 10}}} \over {14 \times {{10}^{ - 10}}}}$

= 0.059 $\times$ ($-$ 0.6274) = $-$ 0.037 V

The negative value of E_cell indicates that the cell reaction $[AgBr(s) + C{l^ - }(aq) \to AgCl(s) + B{r^ - }(aq)]$ is not spontaneous but the reverse reaction is spontaneous as in this case E_cell = +0.037 V. To carry out the reverse reaction, the polarity of the given cell should be reversed i.e., the anode should be made cathode, and vice-versa.

Since $E_{Z{n^{2 + }}|Zn}^0 < E_{N{i^{2 + }}|Ni}^0$, Zn will undergo oxidation and Ni²⁺ reduction.

The reaction that occurs due to addition of Zn-granules to the solution of nickel nitrate is :

$Zn(s) + N{i^{2 + }}(aq) \to Z{n^{2 + }}(aq) + Ni(s)$ ....... (1)

Suppose, the concentration of Zn²⁺ ions at equilibrium = xM. Hence, at equilibrium concentration of Ni²⁺ = (1 $-$ x) M [$\because$ initial concentration of Ni²⁺ = 1 M and 1 mol of Zn reacts with 1 mol of Ni²⁺]

$\therefore$ [Zn²⁺] = x M and [Ni²⁺] = (1 $-$ x) M

Nernst equation for the reaction (1) is :

${E_{cell}} = E_{cell}^0 - {{0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {[N{i^{2 + }}]}}$

$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = [ - 0.24 - ( - 0.75)]V = 0.51V$

At equilibrium, E_cell = 0 $\therefore$ $0 = 0.51 - {{0.059} \over 2}\log {x \over {1 - x}}$

or, $\log {x \over {1 - x}} = 17.28$ or, ${x \over {1 - x}} = 1.9 \times {10^{17}}$

or, ${1 \over {1 - x}} - 1 = 1.9 \times {10^{17}}$ $\therefore$ $(1 - x) = 5.26 \times {10^{ - 18}}$

$\therefore$ Concentration of Ni²⁺ at equilibrium = 5.26 $\times$ 10^$-$18 M

To solve this problem, we need to calculate the amount of Zn²⁺ ions that get deposited during the electrolysis and then use this information to find the final molarity of Zn²⁺ ions in the solution.

First, let's calculate the total charge passed through the cell:

$ Q = I \times t $

$ Q = 1.70 \, \text{A} \times 230 \, \text{s} $

$ Q = 391 \, \text{C} $

Next, we consider the current efficiency, which is 90%. Therefore, the effective charge used for the deposition of zinc is:

$ Q_{\text{effective}} = 0.90 \times 391 \, \text{C} $

$ Q_{\text{effective}} = 351.9 \, \text{C} $

Now, we use Faraday's laws of electrolysis to find the amount of zinc deposited. The molar mass of zinc (Zn) is 65.38 g/mol, and the charge per mole of Zn²⁺ (as it releases two electrons per ion) is:

$ n_{\text{electrons}} = 2 \times F $

$ n_{\text{electrons}} = 2 \times 96485 \, \text{C/mol} $

$ n_{\text{electrons}} = 192970 \, \text{C/mol} $

Using the effective charge, the moles of Zn deposited can be calculated as:

$ \text{moles of Zn} = \frac{Q_{\text{effective}}}{n_{\text{electrons}}} $

$ \text{moles of Zn} = \frac{351.9 \, \text{C}}{192970 \, \text{C/mol}} $

$ \text{moles of Zn} = 0.001824 \, \text{mol} $

The initial moles of Zn²⁺ in the solution can be calculated from its initial concentration and volume:

$ \text{moles of Zn}\,^{2+}_{\text{initial}} = M \times V $

$ \text{moles of Zn}\,^{2+}_{\text{initial}} = 0.160 \, \text{M} \times 0.300 \, \text{L} $

$ \text{moles of Zn}\,^{2+}_{\text{initial}} = 0.048 \, \text{mol} $

After deposition, the moles of Zn²⁺ decreases by the moles of Zn deposited:

$ \text{moles of Zn}\,^{2+}_{\text{final}} = \text{moles of Zn}\,^{2+}_{\text{initial}} - \text{moles of Zn} $

$ \text{moles of Zn}\,^{2+}_{\text{final}} = 0.048 \, \text{mol} - 0.001824 \, \text{mol} $

$ \text{moles of Zn}\,^{2+}_{\text{final}} = 0.046176 \, \text{mol} $

Finally, the final concentration of Zn²⁺ ions can be calculated maintaining the same volume:

$ C_{\text{final}} = \frac{\text{moles of Zn}\,^{2+}_{\text{final}}}{V} $

$ C_{\text{final}} = \frac{0.046176 \, \text{mol}}{0.300 \, \text{L}} $

$ C_{\text{final}} = 0.1539 \, \text{M} $

Thus, the final molarity of Zn²⁺ in the solution after 230 seconds of electrolysis with a current of 1.70 A and a current efficiency of 90% is approximately 0.154 M.

To answer these questions, we first need to understand two key concepts:

• Normality (N), which differs from molarity and relates to the number of equivalents per liter of solution.
• The equivalent weight of the compound, which in this case involves understanding how many electrons are transferred in the half-cell reaction.

Part (i): Let's calculate based on the half-cell reaction:

$BrO_3^- + 6H^+ + 6e^- \rightarrow Br^- + 3H_2O$

This reaction shows each mole of $BrO_3^-$ consumes 6 equivalents (6 moles of electrons). The molar mass of sodium bromate ($NaBrO_3$) is $22.99 + 79.904 + 48.00 = 150.894\, g/mol$.

To find the weight of sodium bromate needed:

1. The equivalent weight of $NaBrO_3$ = $\frac{Molar\, mass}{6} = \frac{150.894}{6} = 25.149\, g/equiv$.

2. To make a 0.672 N solution in 85.5 mL, first convert mL to L: $85.5\, mL = 0.0855\, L$.

3. Calculate the total equivalents needed: $0.672\, eq/L \times 0.0855\, L = 0.05748\, equivalents$.

4. Calculate the mass of $NaBrO_3$ required: $0.05748\, equivalents \times 25.149\, g/equiv = 1.446\, g$.

To find the molarity (M) of the solution:

The molarity would be the number of moles of $NaBrO_3$ in the given volume of solution:

1. Number of moles = $\frac{1.446\, g}{150.894\, g/mol} = 0.00958\, moles$.

2. Molarity = $\frac{0.00958\, moles}{0.0855\, L} = 0.112\, M$.

Part (ii): Consider the revised half-cell reaction:

$2BrO_3^- + 12H^+ + 10e^- \rightarrow Br_2 + 6H_2O$

Here, 2 moles of $BrO_3^-$ are reduced by 10 electrons. Each mole of $BrO_3^-$ consumes 5 equivalents (5 electrons):

1. The equivalent weight of $NaBrO_3$ in this case = $\frac{Molar\, mass}{5} = \frac{150.894}{5} = 30.179\, g/equiv$.

Calculating the mass required to produce a 0.672 N solution:

1. Total equivalents required: $0.672\, eq/L \times 0.0855\, L = 0.05748\, equivalents$

2. Mass of $NaBrO_3$ required: $0.05748\, equivalents \times 30.179\, g/equiv = 1.734\, g$.

Calculating the molarity based on these equivalents:

1. Number of moles = $\frac{1.734\, g}{150.894\, g/mol} = 0.0115\, moles$.

2. Molarity = $\frac{0.0115\, moles}{0.0855\, L} = 0.134\, M$.

In summary, the weight and molarity needed under each half-cell reaction are calculated by considering the number of electrons transferred per mole of reactant and adjusting the equivalent weight accordingly.