An aqueous solution containing Na⁺, Sn²⁺, Cl^– and SO₄^2– ions, all at unit concentration, is electrolyzed between a silver anode and a platinum cathode. What changes occur at the electrodes when current is passed through the cell? Given: \(E^0_{Ag^+|Ag}= 0.799 V\)

\(E^0_{Sn^{2+}|Sn}= -0.14V, E^0_{Cl_2|Cl^-}\)

\(E^0_{S_2O_8^{2-}|SO_4^{2-}}=2V, E^0_{Sn^{4+}|Sn^{2+}}= 0.13 V\)

Sn²⁺ is reduced and Sn²⁺ is oxidised

At anode either Ag can be oxidized to \(Ag^+\) or \(Sn^{2+}\) to \(Sn^{4+}\) or \(Cl^+\) to \(Cl_2\) or \(SO_4^{2-}\) to \(S_2O_8^{2-}\). Their respective oxidation potential values are \(-0.799V\), \(0.13V\), \(-1.36V\) and \(-2V\). From these values it is evident that \(Sn^{2+}\) would be oxidized first, followed by \(Ag\) at anode. At cathode, either \(Na^+\) can get reduced to \(Na\) or \(Sn^{2+}\) or \(Sn^{2+}\) to \(H^+\) or \(H_2\). The reduction potential value for \(Na^+\) is highly negative while for \(Sn^{2+} | Sn\) is \(-0.14V\) and for \(H^+ + e^- \rightarrow \frac{1}{2}H_2\left(E_{H^+|H_2} = 0.059\frac{log(1)}{10^{-7}}\right)\) is \(-0.413V\). Thus \(Sn^{2+}\) will get reduced at cathode followed by \(H^+\).

Therefore, the answer is (3) Sn²⁺ is reduced and Sn²⁺ is oxidized.