Question
Using Nernst equation, calculate the potentials for the following half reactions :
(a) $I _{2( s )}+2 e ^{-} \longrightarrow 2 I ^{-},\left[ I ^{-}\right]=0.03 M$,
$E_{ I ^{-} / I _2}^0= 0 . 5 3 5 V .$
(b) $Fe _{\text {(aq) }}^{3+}+ e ^{-} \longrightarrow Fe _{(\text {aq }}^{2+},\left[ Fe ^{2+}\right]=0.1 M$, $\left[ Fe ^{3+}\right]=0.01 M \quad E ^0 Fe ^{3+}, Fe ^{2+}=0.771 V$.

Answer

(a) Given:
$E_{ I ^{-} / I _2}^0=0.535 V ;\left[ I ^{-}\right]=0.03 M$
Half cell is, $I _{( aq )}^{-}\left| I _{2( s )}\right| Pt$
Reduction reaction $: I _{2( s )}+2 e ^{-} \longrightarrow 2 I _{( aq )}^{-}$
$\therefore n=2 \quad$
$\begin{aligned}
& E_{ I _2 / I^{-}}=E_{ I _2 / I ^{-}}^0-\frac{0.0592}{n} \log _{10} \frac{\left[ I ^{-}\right]^2}{\left[ I _2 I _{ S }\right.} \\
&=0.535-\frac{0.0592}{2} \log _{10} \frac{(0.03)^2}{1} \\
&=0.535-0.0296 \log _{10} 9 \times 10^{-4} \\
&=0.535-0.0296[4.9542] \\
&=0.535-0.0296[-4+0.9542] \\
&=0.535-0.0296[-3.0458] \\
&= 0.535+0.0902 \\
&=0.6252 V
\end{aligned}$

(b) Given: $E^0 Fe ^{3+}, Fe ^{2+}=0.771 V ;\left[ Fe ^{2+}\right]=0.1 M$,
$\left[ Fe ^{3+}\right]=0.01 M$
Half cell is, $Fe _{( aq )}^{2+}, Fe _{( aq )}^{3+} \mid Pt$
Reduction reaction : $Fe _{( aq )}^{3+}+ e ^{-} \longrightarrow Fe _{( aq )}^{2+}$
$\therefore n=1 \quad $
$\begin{aligned}
E_{ Fe ^{3+}, Fe ^{2+}} & =E^0 Fe ^{3+}, Fe ^{2+}-\frac{0.0592}{n} \log _{10} \frac{\left[ Fe ^{2+}\right]}{\left[ Fe ^{3+}\right]} \\
& =0.771-\frac{0.0592}{1} \log _{10} \frac{0.1}{0.01} \\
& =0.771-0.0592 \log _{10} 10 \\
& =0.771-0.0592 \times 1 \\
& =0.7118 V
\end{aligned}$
(a) Potential of the half cell $=0.6252 V$
(b) Potential of the half cell $=0.7118 V$.
(a) Given:
$E_{ I ^{-} / I _2}^0=0.535 V ;\left[ I ^{-}\right]=0.03 M$
Half cell is, $I _{( aq )}^{-}\left| I _{2( s )}\right| Pt$
Reduction reaction $: I _{2( s )}+2 e ^{-} \longrightarrow 2 I _{( aq )}^{-}$
$\therefore n=2 \quad$
$\begin{aligned}
& E_{ I _2 / I^{-}}=E_{ I _2 / I ^{-}}^0-\frac{0.0592}{n} \log _{10} \frac{\left[ I ^{-}\right]^2}{\left[ I _2 I _{ S }\right.} \\
&=0.535-\frac{0.0592}{2} \log _{10} \frac{(0.03)^2}{1} \\
&=0.535-0.0296 \log _{10} 9 \times 10^{-4} \\
&=0.535-0.0296[4.9542] \\
&=0.535-0.0296[-4+0.9542] \\
&=0.535-0.0296[-3.0458] \\
&= 0.535+0.0902 \\
&=0.6252 V
\end{aligned}$

(b) Given: $E^0 Fe ^{3+}, Fe ^{2+}=0.771 V ;\left[ Fe ^{2+}\right]=0.1 M$,
$\left[ Fe ^{3+}\right]=0.01 M$
Half cell is, $Fe _{( aq )}^{2+}, Fe _{( aq )}^{3+} \mid Pt$
Reduction reaction : $Fe _{( aq )}^{3+}+ e ^{-} \longrightarrow Fe _{( aq )}^{2+}$
$\therefore n=1 \quad $
$\begin{aligned}
E_{ Fe ^{3+}, Fe ^{2+}} & =E^0 Fe ^{3+}, Fe ^{2+}-\frac{0.0592}{n} \log _{10} \frac{\left[ Fe ^{2+}\right]}{\left[ Fe ^{3+}\right]} \\
& =0.771-\frac{0.0592}{1} \log _{10} \frac{0.1}{0.01} \\
& =0.771-0.0592 \log _{10} 10 \\
& =0.771-0.0592 \times 1 \\
& =0.7118 V
\end{aligned}$
(a) Potential of the half cell $=0.6252 V$
(b) Potential of the half cell $=0.7118 V$.

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