What is the EMF of the following cell at 298 K?
Fe(s) | Fe2+ (0.001 M) || H+ (1M) | H2(g) (1 bar), Pt(s) (Given E°cell = +0.44V)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.529 V
The emf of a cell at concentrations other than 1 M is given by the Nernst equation. At 298 K:
\( \displaystyle E_{\text{cell}}=E^{\circ}_{\text{cell}}-\frac{0.059\,\mathrm{V}}{n}\log\frac{[\text{products}]}{[\text{reactants}]} \)
The left electrode (iron) is the anode and the hydrogen electrode on the right is the cathode:
Anode: \( \displaystyle \ce{Fe(s) -> Fe^2+(aq) + 2e^-} \)
Cathode: \( \displaystyle \ce{2H^+(aq) + 2e^- -> H2(g)} \)
Cell reaction: \( \displaystyle \ce{Fe(s) + 2H^+(aq) -> Fe^2+(aq) + H2(g)} \)
Two electrons are transferred, so n = 2. Iron is a solid and is left out; \( \ce{H2} \) is at 1 bar, so its term is 1.
\( \displaystyle E_{\text{cell}}=E^{\circ}_{\text{cell}}-\frac{0.059}{2}\log\frac{[\ce{Fe^2+}]\,p_{\ce{H2}}}{[\ce{H^+}]^2} \)
\( \displaystyle E_{\text{cell}}=0.44-\frac{0.059}{2}\log\frac{10^{-3}\times1}{(1)^2} \)
\( \displaystyle E_{\text{cell}}=0.44-0.0295\times(-3)=0.44+0.0885 \)
\( \displaystyle E_{\text{cell}}=0.5285\approx0.529\,\mathrm{V} \)
The emf is higher than \( E^{\circ}_{\text{cell}} \) because the product ion \( \ce{Fe^2+} \) is very dilute, so the reaction quotient is less than 1.