Electrochemistry

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.