Question

Consider the cell, $P t \mid H_{2}(g, 1\, atm) H^{+}(a q, 1\, M) \| F e^{3+}(a q)$ $F e^{2+}(a q) \mid P t(s)$ Given that $E^{o}{ }_{F e^{3+} / F e^{2+}}=0.771\, V$, the ratio of concentration of $F e_{(a q)}^{2+}$ to $F e_{(a q)}^{3+}$ is, when the cell potential is $0.830\, V$.

• A. 0.101
• B. 0.924
• C. 0.12
• D. None of these

Answer 1

Answer: (A)

0.101

Answer 2

The half-cell reactions of the given cell are as
At anode $\frac{1}{2} H_{2} \rightarrow H^{+}+e^{-} ; E_{1}^{o}=-0.00\, V$
At cathode $F e^{3+}+e^{-} \rightarrow F e^{2+} ; E_{?}^{o}=0.771\, V$ Net reaction
$Fe ^{3+}+\frac{1}{2} H _{2} \rightarrow Fe ^{2+}+ H ^{+} ; E_{\text {cell }}^{o}=0.771\, V -0.00\, V =0.771\, V$ From Nernst
$E_{\text {cell }}=E_{\text {cell }}^{o}=-\frac{0.0591}{n} \log \frac{\left[ Fe ^{2+}\right]\left[ H ^{+}\right]}{\left[ Fe ^{3+}\right]+ pH _{2}^{1 / 2}}$
$0.830=0.771-\frac{0.0591}{1} \log \left[\frac{F e^{2+}}{F e^{3+}}\right]-\frac{0.059}{0.0591}=\log \frac{\left[F e^{2+}\right]}{\left[F e^{3+}\right]}$
$\therefore \frac{\left[ Fe ^{2+}\right]}{\left[ Fe ^{3+}\right]}=$ antilog
$(-0.998)=0.101$