Cu^{2+} + 2e^- to Cu. On increasing [Cu^{2+}] | vedclass

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(A) The reduction half-reaction is: $Cu^{2+}_{(aq)} + 2e^- 	o Cu_{(s)}$According to the Nernst equation:$E_{Cu^{2+}/Cu} = E^0_{Cu^{2+}/Cu} - \frac{0.

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(A) The reduction half-reaction is: $Cu^{2+}_{(aq)} + 2e^- 	o Cu_{(s)}$According to the Nernst equation:$E_{Cu^{2+}/Cu} = E^0_{Cu^{2+}/Cu} - \frac{0.059}{2} \log \frac{1}{[Cu^{2+}]}$This can be rewritten as:$E_{Cu^{2+}/Cu} = E^0_{Cu^{2+}/Cu} + \frac{0.059}{2} \log [Cu^{2+}]$From the equation,it is clear that $E_{Cu^{2+}/Cu}$ is directly proportional to $\log [Cu^{2+}]$.Therefore,as the concentration $[Cu^{2+}]$ increases,the electrode potential $E_{Cu^{2+}/Cu}$ also increases.

$Cu^{2+} + 2e^- \to Cu$. On increasing $[Cu^{2+}]$ concentration,electrode potential

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