A cell is prepared by dipping a \(Cu\) rod in \(\text{1 M }CuSO_4\) solution and \(Sn\) rod in \(\text{1 M }SnCl_2\) solution. The standard electrode potential of \(Cu\) is \(+0.34 V\) and Sn is \(-0.14 V\). Emf of cell will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(0.48 V\)
In a galvanic cell the electrode with the higher reduction potential acts as the cathode and the other as the anode:
\( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \)
E°(\( \ce{Cu^{2+}} \)/Cu) = +0.34 V is higher than E°(\( \ce{Sn^{2+}} \)/Sn) = −0.14 V, so copper is the cathode and tin the anode:
Anode: \( \ce{Sn(s) -> Sn^2+(aq) + 2e-} \)
Cathode: \( \ce{Cu^2+(aq) + 2e- -> Cu(s)} \)
Cell: \( \ce{Sn(s) + Cu^2+(aq) -> Sn^2+(aq) + Cu(s)} \)
Both solutions are 1 M, so the emf equals the standard emf:
\( E^\circ_{cell} = 0.34 - (-0.14) = 0.48\ \text{V} \)