The value of $E^{\circ}_{cell}$ for the following reaction is:
$Cu^{2+} + Sn^{2+} \rightarrow Cu + Sn^{4+}$
(G

Question

The value of $E^{\circ}_{cell}$ for the following reaction is:
$Cu^{2+} + Sn^{2+} \rightarrow Cu + Sn^{4+}$
(Given, equilibrium constant is $10^6$)

Accepted Answer

The relationship between standard cell potential ($E^{\circ}_{cell}$) and the equilibrium constant ($K_c$) at $298	ext{ K}$ is given by the equation:
$E^{\circ}_{cell} = \frac{0.0591}{n} \log K_c$
For the given cell reaction:
Oxidation half-reaction: $Sn^{2+} ightarrow Sn^{4+} + 2e^-$
Reduction half-reaction: $Cu^{2+} + 2e^- ightarrow Cu$
The number of electrons transferred ($n$) is $2$.
Given $K_c = 10^6$,
$E^{\circ}_{cell} = \frac{0.0591}{2} \log(10^6)$
$E^{\circ}_{cell} = \frac{0.0591}{2} 	imes 6 = 0.0591 	imes 3 = 0.1773	ext{ V} \approx 0.17	ext{ V}$

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