The equilibrium constant for the reaction is at 298 K. The magnitude of standard electrode potential of is ______. (Nearest integer) [2023]

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Solution

Q.
The equilibrium constant for the reaction

$Zn (s) + {Sn}^{2 +} (aq) ⇌ {Zn}^{2 +} (aq) + Sn (s)$ is $1 \times 10^{20}$ at 298 K. The magnitude of standard electrode potential of

$Sn / {Sn}^{2 +} if E_{{Zn}^{2 +} / Zn}^{θ} = - 0.76 V$ is ______ $\times 10^{- 2} V$ . (Nearest integer)

$Given: \frac{2.303 RT}{F} = 0.059 V$ [2023]

Ans.
(17)

$E ° = \frac{0.059}{2} \log K_{eq}$

$= \frac{0.059}{2} \log 10^{20} = \frac{0.059}{2} \times 20 = 0.59 V$

$E_{cell}^{\circ} = E_{S n^{2 +} / S n}^{\circ} - E_{Z n^{2 +} / Z n}^{\circ}$

$0.59 = E_{S n^{2 +} / S n}^{\circ} - (- 0.76)$

$E_{S n^{2 +} / S n}^{\circ} = - 0.76 + 0.59 = - 0.17$

$E_{S n^{2 +} / S n}^{\circ} = - 0.76 + 0.59 = 17 \times 10^{- 2} = 17$

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