Question

Mark the correct Nernst equation for the given cell.
$F{e}_{\left(s\right)}\left|F{e}^{2+}\left(0.001M\right)\right|\left|H^{+}\left(1M\right)\right|H_{2\left(g\right)}\left(1bar\right)|P{t}_{\left(s\right)}$

• A. $E_{cell}=E_{cell}^{\circ }-\frac{0.591}{2}log\frac{\left[F{e}^{2+}\right]{\left[H^{+}\right]}^2}{\left[Fe\right]\left[H_2\right]}$
• B. $E_{cell}=E_{cell}^{\circ }-\frac{0.591}{2}log\frac{\left[Fe\right]{\left[H^{+}\right]}^2}{\left[F{e}^{2+}\right]\left[H_2\right]}$
• C. $E_{cell}=E_{cell}^{\circ }-\frac{0.591}{2}log\frac{\left[F{e}^{2+}\right]\left[H_2\right]}{\left[Fe\right]{\left[H^{+}\right]}^2}$
• D. $E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}log\frac{\left[Fe\right]\left[H_2\right]}{\left[F{e}^{2+}\right]{\left[H^{+}\right]}^2}$

Answer 1

Answer: (C)

At anode : $Fe\to F{e}^{2+}\left(0.001M\right)+2e^{-}$
At cathode : $2H^{+}\left(1M\right)+2e^{-}\to H_2\left(1atm\right)$
Net reaction : $Fe+2H^{+}\to F{e}^{2+}+H_2$
Nernst equation for the given cell,
$E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}log\frac{\left[F{e}^{2+}\right]\left[H_2\right]}{\left[Fe\right]{\left[H^{+}\right]}^2}$