Question

Calculate the reduction potential of a half-cell containing of platinum electrode immersed in
$2.0MF{e}^{2+}$ and $0.02MF{e}^{3+}.$
$F{e}^{3+}+e^{-}\to F{e}^{2+}$
Given : $E_{F{e}^{3+}/F{e}^{2+}}^o=0.771V;F{e}^{3+}+e^{-}\to F{e}^{2+}$

• A. 0.653 V
• B. 0.889 V
• C. 0.0653 V
• D. 2.771 V

Answer 1

Answer: (A)

The Nernst's equation, for the half-cell reaction,
$F{e}^{3+}+e^{-}\to F{e}^{2+}$
is $E_{F{e}^{3+}/F{e}^{2+}}=E_{F{e}^{3+}/F{e}^{2+}}^o-\frac{0.0591}{1}\log \frac{\left[F{e}^{2+}\right]}{\left[F{e}^{3+}\right]}$
$=0.771-0.0591\log \frac{2.00}{0.02}$
$=0.771-0.0591\log 10^2$
$=0.771-0.0591\times 2$
$=+0.653V$