For a given half-cell,Al^{3+} + 3e^{-} righta | vedclass

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(D) The half-cell reaction is $Al^{3+} + 3e^{-} \rightarrow Al_{(s)}$.According to the Nernst equation:$E_{Red} = E_{Red}^{o} - \frac{0.0591}{3} \log

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(D) The half-cell reaction is $Al^{3+} + 3e^{-} \rightarrow Al_{(s)}$.According to the Nernst equation:$E_{Red} = E_{Red}^{o} - \frac{0.0591}{3} \log \frac{1}{[Al^{3+}]}$ (since the active mass of solid $Al = 1$).This simplifies to: $E_{Red} = E_{Red}^{o} + \frac{0.0591}{3} \log [Al^{3+}]$.From this expression,it is clear that $E_{Red}$ is directly proportional to $\log [Al^{3+}]$.Therefore,as the concentration of $Al^{3+}$ ions increases,the electrode potential $E_{Red}$ will increase.

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

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