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