Question

Mark the correct Nernst equation for the given cell.
$Fe_{\left(s\right)}\left|Fe^{2+}\left(0.001\,M\right)\right|\left|H^{+}\left(1\,M\right)\right|H_{2\left(g\right)}\,\left(1\,bar\right)| Pt_{\left(s\right)}$

• A. $E_{cell}=E^{\circ}_{cell}-\frac{0.591}{2}log \frac{\left[Fe^{2+}\right]\left[H^{+}\right]^{2}}{\left[Fe\right]\left[H_{2}\right]}$
• B. $E_{cell}=E^{\circ}_{cell}-\frac{0.591}{2}log \frac{\left[Fe\right]\left[H^{+}\right]^{2}}{\left[Fe^{2+}\right]\left[H_{2}\right]}$
• C. $E_{cell}=E^{\circ}_{cell}-\frac{0.591}{2}log \frac{\left[Fe^{2+}\right]\left[H_{2}\right]}{\left[Fe\right]\left[H^{+}\right]^{2}}$
• D. $E_{cell}=E^{\circ}_{cell}-\frac{0.0591}{2}log \frac{\left[Fe\right]\left[H_{2}\right]}{\left[Fe^{2+}\right]\left[H^{+}\right]^{2}}$

Answer 1

Answer: (C)

At anode : $Fe \to Fe^{2+}\left(0.001\,M\right)+2e^{-}$
At cathode : $2H^{+}\left(1\,M\right)+2e^{-} \to H_{2}\left(1\,atm\right)$
Net reaction : $Fe+2H^{+} \to Fe^{2+}+H_{2}$
Nernst equation for the given cell,
$E_{cell}=E^{\circ}_{cell}-\frac{0.0591}{2} log \frac{\left[Fe^{2+}\right]\left[H_{2}\right]}{\left[Fe\right]\left[H^{+}\right]^{2}}$