KCET · Chemistry · Electrochemistry
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
- A \(P>S>R>Q\)
- B \(S>R>Q < P\)
- C \(Q>R>S>P\)
- D \(P>Q>R>S\)
Answer & Solution
Correct Answer
(C) \(Q>R>S>P\)
Step-by-step Solution
Detailed explanation
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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