Unit 3 Electrochemistry (Intext Questions-2)
Intext Questions
3.4 Calculate the potential of hydrogen electrode in contact with a solution whose $\mathrm{pH}$ is 10 .
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Answer
For hydrogen electrode,$\mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H_2} \text {, it is given that } \mathrm{pH}=10$
$\therefore\left[\mathrm{H}^{+}\right]=10^{-10} \mathrm{M}$
Now, using Nernst equation:
$ \mathrm{H_{(\mathrm{H}^{+} / \frac{1}{2} \mathrm{H_2} )}}=E_{(\mathrm{H}^{+} / \frac{1}{2} \mathrm{H_2} )}^{\ominus}-\frac{\mathrm{R} T}{n \mathrm{~F}} \ln \frac{1}{ [\mathrm{H}^{+} ]}$
$ \begin{aligned} =E_{(\mathrm{H}^{+} / \frac{1}{2} \mathrm{H_2})}^{\ominus}-\frac{0.0591}{1} \log \frac{1}{[\mathrm{H}^{+}]} \\ =0-\frac{0.0591}{1} \log \frac{1}{[10^{-10}]} \\ =-0.0591 \log 10^{10} \\ =-0.591 \mathrm{~V} \end{aligned} $
3.5 Calculate the emf of the cell in which the following reaction takes place: $\mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.002 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(0.160 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s})$ Given that $E_{\text {cell }}^{o}=1.05 \mathrm{~V}$
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Answer
Applying Nernst equation we have:
$$ \begin{aligned} & E_{\text {(cell) }}=E_{\text {(cell) }}^{\ominus}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Ni}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}} \\ \\ & =1.05-\frac{0.0591}{2} \log \frac{(0.160)}{(0.002)^{2}} \\ \\ & =1.05-0.02955 \log \frac{0.16}{0.000004} \\ \\ & =1.05-0.02955 \log 4 \times 10^{4} \\ \\ & =1.05-0.02955(\log 10000+\log 4) \\ \\ & =1.05-0.02955(4+0.6021) \\ \\ & =0.914 \mathrm{~V} \end{aligned} $$
3.6 The cell in which the following reaction occurs: $ 2 \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{I_2}(\mathrm{~s})$ has $E_{\text {cell }}^{\mathrm{o}}=0.236 \mathrm{~V}$ at $298 \mathrm{~K}$. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
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Answer
Here, $n=2, E_{\text {cell }}^{\ominus}=0.236 \mathrm{~V},{ _\mathrm{T}}=298 \mathrm{~K}$
We know that:
$\Delta_{r} \mathrm{G}^{\ominus}=-n \mathrm{FE_\text {cell }}^{\ominus}$
$=-2 \times 96487 \times 0.236$
$=-45541.864 \mathrm{~J} \mathrm{~mol}^{-1}$
$=-45.54 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Again, $\Delta_r G^{\ominus}= -2.303 R T \log K_\mathrm{c}$
$\Rightarrow \log K_{\mathrm{c}}=-\frac{\Delta_{r} G^{\ominus}}{2.303 \mathrm{R} T}$
$ =-\frac{-45.54 \times 10^{3}}{2.303 \times 8.314 \times 298} $
$=7.981$
$\therefore K_{\mathrm{c}}=$ Antilog (7.981)
$=9.57 \times 10^{7}$
