If the standard electrode potential, \( E_{cell}^o = 1.10 V \), then the \( E_{cell} \) for the following reaction at 298 K is
\( Zn(s) + Cu^{2+} (0.1 M) →Zn^{2+} (0.01M) + Cu(s) \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.1295 V
The emf of a cell at concentrations other than 1 M is given by the Nernst equation. At 298 K:
\( \displaystyle E_{cell} = E^\circ_{cell} - \frac{0.059}{n}\log \frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]} \)
The cell reaction \( \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} \) transfers 2 electrons (Zn goes from 0 to +2, Cu from +2 to 0), so n = 2. The solids do not appear in the quotient.
\( \displaystyle E_{cell} = 1.10 - \frac{0.059}{2}\log \frac{0.01}{0.1} \)
\( \displaystyle E_{cell} = 1.10 - 0.0295 \times \log 10^{-1} = 1.10 - 0.0295 \times (-1) \)
\( \displaystyle E_{cell} = 1.10 + 0.0295 = 1.1295\,\mathrm{V} \)
So the emf of the cell is 1.1295 V.