JEE MAIN - Chemistry (2022 - 29th July Evening Shift - No. 18)
For a cell, $$\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(0.001 \,\mathrm{M}) \| \mathrm{Ag}^{+}(0.01 \,\mathrm{M})\right| \mathrm{Ag}(\mathrm{s})$$
the cell potential is found to be $$0.43 \mathrm{~V}$$ at $$298 \mathrm{~K}$$. The magnitude of standard electrode potential for $$\mathrm{Cu}^{2+} / \mathrm{Cu}$$ is _________ $$\times 10^{-2} \mathrm{~V}$$.
[Given : $$E_{A{g^ + }/Ag}^\Theta $$ = 0.80 V and $${{2.303RT} \over F}$$ = 0.06 V]
Answer
34
Explanation
Anode : $$\mathrm{Cu}(\mathrm{s}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}$$
Cathode : $$\left.[\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(\mathrm{s})\right] 2$$
Cus(s) $$+2 \mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$$
$$ \begin{aligned} & E_{\text {cell }}=E_{\text {cell }}^{0}-\frac{0.06}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}} \\ & 0.43=\mathrm{E}_{\text {cell }}^{0}-\frac{0.06}{2} \log \left(\frac{10^{-3}}{\left(10^{-2}\right)^{2}}\right) \\ & 0.43=E_{\text {cell }}^{0}-0.03 \log 10 \\ & \mathrm{E}_{\text {cell }}^{0}=0.46 \mathrm{~V} \\ & \mathrm{E}_{\text {cell }}^{0}=\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}-\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0} \\ & \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=(0.80-0.46)=0.34 \mathrm{~V}=34 \times 10^{-2} \end{aligned} $$
Cathode : $$\left.[\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(\mathrm{s})\right] 2$$
Cus(s) $$+2 \mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$$
$$ \begin{aligned} & E_{\text {cell }}=E_{\text {cell }}^{0}-\frac{0.06}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}} \\ & 0.43=\mathrm{E}_{\text {cell }}^{0}-\frac{0.06}{2} \log \left(\frac{10^{-3}}{\left(10^{-2}\right)^{2}}\right) \\ & 0.43=E_{\text {cell }}^{0}-0.03 \log 10 \\ & \mathrm{E}_{\text {cell }}^{0}=0.46 \mathrm{~V} \\ & \mathrm{E}_{\text {cell }}^{0}=\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}-\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0} \\ & \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=(0.80-0.46)=0.34 \mathrm{~V}=34 \times 10^{-2} \end{aligned} $$
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