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Consider the following electrodes $P=\mathrm{Zn}^{2+}(0.0001 \mathrm{M}) / \mathrm{Zn}, Q=\mathrm{Zn}^{2+}(0.1$ M) $/ \mathrm{Zn}$ $R=\mathrm{Zn}^{2+}(0.01 \mathrm{M}) / \mathrm{Zn}, S=\mathrm{Zn}^{2+}(0.001 \mathrm{M}) / \mathrm{Zn}$ $E^{\circ}\left(\mathrm{Zn} / \mathrm{Zn}^{2+}\right)=-0.76 \mathrm{~V}$ electrode potentials of the above electrodes in volts are in the order
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Verified Answer
The correct answer is:
$Q>R>S>P$
The standard potential of $\mathrm{Zn} \mid \mathrm{Zn}^{2+}$ half-cell $=-0.76 \mathrm{~V}$
Half-cell reaction,
$\overline{\mathrm{Z}} \mathrm{n}^{2 *}(a q)+\overline{\mathcal{2}} e^{=} \longrightarrow \overline{\mathrm{Z}}(s)$
$E_{\text {red }}=E_{\text {red }}^{\circ}-\frac{0.059}{n} \log \left[\frac{1}{\mathrm{Zn}^{2+}}\right]$
$=-0.76+\frac{0.059}{n} \log \left[\mathrm{Zn}^{2+}\right]$
Lower the concentration of $\mathrm{Zn}^{2+}$, lower is the $E_{\text {red }}$ or vice-versa.
Hence, the correct order is $Q>R>S>P$.
Half-cell reaction,
$\overline{\mathrm{Z}} \mathrm{n}^{2 *}(a q)+\overline{\mathcal{2}} e^{=} \longrightarrow \overline{\mathrm{Z}}(s)$
$E_{\text {red }}=E_{\text {red }}^{\circ}-\frac{0.059}{n} \log \left[\frac{1}{\mathrm{Zn}^{2+}}\right]$
$=-0.76+\frac{0.059}{n} \log \left[\mathrm{Zn}^{2+}\right]$
Lower the concentration of $\mathrm{Zn}^{2+}$, lower is the $E_{\text {red }}$ or vice-versa.
Hence, the correct order is $Q>R>S>P$.
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