What is the reduction electrode potential (in | vedclass

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(B) The reduction reaction for the copper electrode is: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.Using the Nernst equation at $25^{\circ} C$ $(298 \ K)$

Vedclass · Accepted Answer

$0.2809$

Vedclass · Answer

(B) The reduction reaction for the copper electrode is: $Cu^{2+}(aq) + 2e^- ightarrow Cu(s)$.Using the Nernst equation at $25^{\circ} C$ $(298 \ K)$:$E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}$$E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} + \frac{0.0591}{2} \log [Cu^{2+}]$Substituting the given values:$E_{Cu^{2+}/Cu} = 0.34 + \frac{0.0591}{2} \log (10^{-2})$$E_{Cu^{2+}/Cu} = 0.34 + \frac{0.0591}{2} 	imes (-2)$$E_{Cu^{2+}/Cu} = 0.34 - 0.0591$$E_{Cu^{2+}/Cu} = 0.2809 \ V$

What is the reduction electrode potential (in volts) of a copper electrode when $[Cu^{2+}] = 0.01 \ M$ is in a solution at $25^{\circ} C$? ($E^{\circ}$ of $Cu^{2+}/Cu$ electrode is $+0.34 \ V$)

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