Q51 / 180 · KCET 2025 · Chemistry

For a given half cell, $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}$ on increasing of aluminium ion, the electrode potential will

  • A
    Decrease
  • B
    No change
  • C
    First increase then decrease
  • Correct answer
    D
    Increase

Solution

$\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{s})$ $\begin{aligned} & \mathrm{E}_{\text {Red }}=\mathrm{E}_{\text {Red }}^{\mathrm{o}}-\frac{0.0591}{3} \log \frac{[\mathrm{Al}(\mathrm{~s})]}{\left[\mathrm{Al}^{3+}\right]} \\ & \mathrm{E}_{\text {Red }}=\mathrm{E}_{\text {Red }}^{\mathrm{o}}-\frac{0.0591}{4} \log \frac{1}{\left[\mathrm{Al}^{3+}\right]}[\text { Since active mass of solid }=1] \\ & \mathrm{E}_{\text {Red }}=\mathrm{E}_{\text {Red }}^{\mathrm{o}}+\frac{0.0591}{3} \log \left[\mathrm{Al}^{3+}\right] \end{aligned}$ So $E_{\text {Red }} \propto\left[\mathrm{Al}^{3+}\right] \propto \operatorname{conc}^{\mathrm{n}}$ of $\mathrm{Al}^{3+}$

Hence, the correct answer is option: 'D'.

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