Question:
The electrode potentials for
$Cu^{2+}(aq)+e^- \rightarrow Cu^+_{(aq)} $ and $Cu^{+}(aq)+e^- \rightarrow Cu_{(s)} $ are $+ 0.15 \,V$ and $+ 0.50\, V$ respectively.
The value of $E^0_{Cu^{2+}/Cu} $ will be
The electrode potentials for
$Cu^{2+}(aq)+e^- \rightarrow Cu^+_{(aq)} $ and $Cu^{+}(aq)+e^- \rightarrow Cu_{(s)} $ are $+ 0.15 \,V$ and $+ 0.50\, V$ respectively.
The value of $E^0_{Cu^{2+}/Cu} $ will be
Updated On: Jul 29, 2022
- 0.325 V
- 0.650 V
- 0.150 V
- 0.500 V
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The Correct Option is A
Solution and Explanation
$\Delta G _{3}=\Delta G _{1}+\Delta G _{2}$
$\Rightarrow-2\,FE ^{ o }=-1 F \times 0.15+(-1 F \times 0.50)$
$\Rightarrow-2 \,FE ^{ o }=-0.15\, F -0.50\, F$
$\Rightarrow-2 \,FE ^{ o }=- F (0.15+0.50)$
$\therefore E ^{ o }=\frac{0.65}{2}=0.325\, volt$
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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.
