Question:
The electrode potentials for ${Cu^{2+}(aq)+e^-->Cu^+(aq)}$ and ${Cu^{+}(aq)+e^-->Cu(s)}$ are $+ 0.15\, V$ and $+ 0.50 \,V$, respectively. The value of $E^?_{Cu^{2+}/Cu}$ will be :
The electrode potentials for ${Cu^{2+}(aq)+e^-->Cu^+(aq)}$ and ${Cu^{+}(aq)+e^-->Cu(s)}$ are $+ 0.15\, V$ and $+ 0.50 \,V$, respectively. The value of $E^?_{Cu^{2+}/Cu}$ will be :
Updated On: Apr 15, 2024
- $0.500 \,V$
- $0352 \,V$
- $0.650 \,V$
- $0.150 \,V$
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The Correct Option is B
Solution and Explanation
$cu^{2+}+le^{-} \rightarrow Cu^{+}\,; \,\Delta G^{\circ}_{1}=-n_{1}E_{1}^{\circ}F]$
$\frac{cu^{+}+le^{-} \rightarrow Cu\,; \,\Delta G^{\circ }_{2}=-n_{2}E_{2}^{\circ }F}{cu^{2+}+2e^{-} \rightarrow Cu\,; \,\Delta G^{\circ}=\Delta G^{\circ }_{1}=\Delta G^{\circ}_{2}}$
$-nE^{?}F=1\,n_{1}\,E^{?}F+\left(-1\right)n_{2}\,E^{?}_{2}F$
$-nE^{?}\,F=-F\left(n_{1}\,E^{?}_{1}+n_{2}\,E^{?}_{2}F\right)$
$E^{V}=\frac{n_{1}E^{?}_{1}+n_{2}E^{?}_{2}}{n}=\frac{0.15\times1+0.50\times1}{2}=0.325$
$\frac{cu^{+}+le^{-} \rightarrow Cu\,; \,\Delta G^{\circ }_{2}=-n_{2}E_{2}^{\circ }F}{cu^{2+}+2e^{-} \rightarrow Cu\,; \,\Delta G^{\circ}=\Delta G^{\circ }_{1}=\Delta G^{\circ}_{2}}$
$-nE^{?}F=1\,n_{1}\,E^{?}F+\left(-1\right)n_{2}\,E^{?}_{2}F$
$-nE^{?}\,F=-F\left(n_{1}\,E^{?}_{1}+n_{2}\,E^{?}_{2}F\right)$
$E^{V}=\frac{n_{1}E^{?}_{1}+n_{2}E^{?}_{2}}{n}=\frac{0.15\times1+0.50\times1}{2}=0.325$
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Concepts Used:
Nernst Equation
This equation relates the equilibrium cell potential (also called the Nernst potential) to its concentration gradient across a membrane. If there is a concentration gradient for the ion across the membrane, an electric potential will form, and if selective ion channels exist the ion can cross the membrane.
