What is the emf of the following cell at 298 K?
Fe(s) | Fe2+(0.6 M) || Sn2+(0.2 M) | Sn(s)
(E°Fe2+/Fe = −0.44 V; E°Sn2+/Sn = −0.14 V)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.286 V
In the cell as written, iron (left) is the anode and tin (right) is the cathode. The cell reaction, with n = 2, is:
\( \displaystyle \ce{Fe(s) + Sn^2+(aq) -> Fe^2+(aq) + Sn(s)} \)
Standard emf of the cell:
\( \displaystyle E^{\circ}_{\text{cell}}=E^{\circ}_{\text{right}}-E^{\circ}_{\text{left}}=-0.14\,\mathrm{V}-(-0.44\,\mathrm{V})=0.30\,\mathrm{V} \)
By the Nernst equation at 298 K:
\( \displaystyle E_{\text{cell}}=E^{\circ}_{\text{cell}}-\frac{0.059}{n}\log\frac{[\ce{Fe^2+}]}{[\ce{Sn^2+}]} \)
\( \displaystyle E_{\text{cell}}=0.30-\frac{0.059}{2}\log\frac{0.6}{0.2}=0.30-0.0295\times0.477 \)
\( \displaystyle E_{\text{cell}}=0.30-0.014=0.286\,\mathrm{V} \)