Question

The equilibrium constant for the reaction
\(Zn(s)+Sn^{2+}(aq)⇌Zn^{2+}(aq)+Sn(s)\, is\, 1×10^{20}\) at \(298 K\). 
The magnitude of standard electrode potential of \(Sn/Sn^{2+}\) if \(E^{\degree}_{ Zn^{2+}/Zn}= –0.76V \) is _______ \(×10^{−2} V\). (Nearest integer).
(Given : \(\frac{2.303RT}{ F}\) \(=0.059 V \))

Answer 1

Correct Answer: 17

The equilibrium constant is related to the electrode potentials: \[ \Delta G^\circ = -nF E^\circ = -2.303RT \log K \] \[ E^\circ_{\text{cell}} = \frac{0.059}{2} \log K \] Substituting \(K = 1 \times 10^{20}\): \[ E^\circ_{\text{cell}} = \frac{0.059}{2} \log (1 \times 10^{20}) = 0.059 \times 10 = 0.59 \, \text{V} \] Using \(E^\circ_{\text{cell}} = E^\circ_{\text{Sn}^{2+}/\text{Sn}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}}\): \[ 0.59 = E^\circ_{\text{Sn}^{2+}/\text{Sn}} - (-0.76) \] \[ E^\circ_{\text{Sn}^{2+}/\text{Sn}} = 0.59 - 0.76 = 0.17 \, \text{V} = 17 \times 10^{-2} \, \text{V} \]