The standard reduction potentials of the Pt | Fe3+, Fe2+ and Pt | Sn4+, Sn2+ electrodes are +0.77 V and +0.15 V respectively at 25 °C. What is the standard EMF of the reaction given below?
Sn4+ + 2Fe2+ → Sn2+ + 2Fe3+
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → −0.62 V
The standard EMF of a cell reaction is the standard reduction potential of the electrode where reduction occurs (cathode) minus that of the electrode where oxidation occurs (anode).
Reduction: \( \ce{Sn^4+ + 2e^- -> Sn^2+} \), so the Sn4+, Sn2+ electrode is the cathode: E° = +0.15 V.
Oxidation: \( \ce{2Fe^2+ -> 2Fe^3+ + 2e^-} \), so the Fe3+, Fe2+ electrode is the anode: E° = +0.77 V.
\( \displaystyle E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}} \)
\( \displaystyle E^\circ_{\text{cell}}=(+0.15\,\mathrm{V})-(+0.77\,\mathrm{V}) \)
\( \displaystyle E^\circ_{\text{cell}}=-0.62\,\mathrm{V} \)
The multiplying factor 2 for Fe is not applied to E°, because an electrode potential does not depend on how many times the half-reaction is written. The negative EMF shows that the reaction as written is not spontaneous; the reverse reaction, in which \( \ce{Fe^3+} \) oxidises \( \ce{Sn^2+} \), is.