(B) The electrode reaction for the copper cathode is: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.According to the Nernst equation for this electrode:$E_{C

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(B) The electrode reaction for the copper cathode is: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.According to the Nernst equation for this electrode:$E_{Cu^{2+}/Cu} = E_{Cu^{2+}/Cu}^{\ominus} - \frac{0.059}{n} \log \frac{1}{[Cu^{2+}]}$.Given $\frac{2.303 \ RT}{F} = 0.06 \ V$,we use $0.06$ instead of $0.059$:$E_{Cu^{2+}/Cu} = 0.34 - \frac{0.06}{2} \log \frac{1}{0.1}$.$E_{Cu^{2+}/Cu} = 0.34 - 0.03 \log(10)$.Since $\log(10) = 1$,we have:$E_{Cu^{2+}/Cu} = 0.34 - 0.03 = 0.31 \ V$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

In a cell,a copper electrode was used as a cathode. What is the electrode potential (in $V$) of the copper electrode dipped in $0.1 \ M \ Cu^{2+}$ solution at $298 \ K$?
$(E_{Cu^{2+}/Cu}^{\ominus} = 0.34 \ V; \frac{2.303 \ RT}{F} = 0.06 \ V)$

AP EAMCET 2025 All AP EAMCET

A
$0.34$
B
$0.31$
C
$0.37$
D
$0.40$

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In a cell,a copper electrode was used as a cathode. What is the electrode potential (in $V$) of the copper electrode dipped in $0.1 \ M \ Cu^{2+}$ solution at $298 \ K$?$(E_{Cu^{2+}/Cu}^{\ominus} = 0.34 \ V; \frac{2.303 \ RT}{F} = 0.06 \ V)$

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In a cell,a copper electrode was used as a cathode. What is the electrode potential (in $V$) of the copper electrode dipped in $0.1 \ M \ Cu^{2+}$ solution at $298 \ K$?
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