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