Question

Mark the correct Nernst equation for the given cell.

$F{e}_{\left(s\right)}|F{e}^{2+}\left(0.001M\right)|H^{+}\left(1M\right)|H_{2\left(g\right)}\left(1bar\right)|P{t}_{\left(s\right)}$ :

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

Answer 1

The correct option is C $E_{cell}=E_{cell}^o-\frac{0.0591}{2}log\frac{\left[F{e}^{2+}\right]\left[H_2\right]}{\left[Fe\right][H^{+}{]}^2}$

At anode:

$Fe\to F{e}^{2+}\left(0.0001M\right)+2e^{-}$

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,

$E_{cell}=E_{cell}^0-\frac{0.0591}{n}log\frac{\left[F{e}^{2+}\right]\left[H_2\right]}{\left[Fe\right][H^{+}{]}^2}$

Hence, the correct answer is option $\text{C}$.