Consider the cell, If the potential of the cell is 0.712 V, the ratio of concentrations of Fe 2 + to Fe 3 + is ________ (Nearest integer). [2023]

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Solution

Q.
Consider the cell

${Pt(s)|H}_{2} (g,1 atm) | H^{+} (aq, [H^{+}] = 1) ‖ {Fe}^{3 +} (aq), {Fe}^{2 +} (aq) | Pt (s)$

$Given: E_{{Fe}^{3 +} / {Fe}^{2 +}}^{\circ} = 0.771 V$ and $E_{H^{+} / 1 / 2 H_{2}}^{\circ} = 0 V, T = 298 K$

If the potential of the cell is 0.712 V, the ratio of concentrations of ${Fe}^{2 +}$ to ${Fe}^{3 +}$ is ________ (Nearest integer). [2023]

Ans.
(10)

${Pt(s)|H}_{2} (g,1 atm) | H^{+} (aq,1 M) ‖ {Fe}^{3 +} (aq), {Fe}^{2 +} (aq) | Pt (s)$

${At anode: H}_{2} \to 2 H^{+} + 2 e^{-}$

$At cathode:$ $({Fe}^{3 +} (a q) + e^{-} \to {Fe}^{2 +} (a q)) \times 2$

overall cell reaction:

$H_{2} (g) + 2 {Fe}^{3 +} \to 2 H^{+} (a q) + 2 {Fe}^{2 +} (a q)$

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.771 - 0 = 0.771 V$

$E_{cell} = E_{cell}^{\circ} - \frac{0.06}{n} \log \frac{[{Fe}^{2 +}] {[H^{+}]}^{2}}{P_{H_{2}} . {[{Fe}^{3 +}]}^{2}}$

$0.712 = 0.771 - \frac{0.06}{2} \log \frac{{[{Fe}^{2 +}]}^{2}}{{[{Fe}^{3 +}]}^{2}}$

$\log \frac{[{Fe}^{2 +}]}{[{Fe}^{3 +}]} = \frac{0.771 - 0.712}{0.06} = 1$

$∴$ $\frac{[{Fe}^{2 +}]}{[{Fe}^{3 +}]} = 10$

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