Question

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

• 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,
$F e^{3+}+e^{-} \rightarrow F e^{2+}$
is $E_{F e^{3+} / F e^{2+}}=E_{F e^{3+} / F e^{2+}}^{o}-\frac{0.0591}{1} \log \frac{\left[F e^{2+}\right]}{\left[F e^{3+}\right]}$
$=0.771-0.0591 \log \frac{2.00}{0.02}$
$=0.771-0.0591 \log 10^{2}$
$=0.771-0.0591 \times 2$
$=+0.653\, V$