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The cell potential of the following cell at $ \text{25}{{\,}^{\text{o}}}\text{C} $ (in volts) is $ \underset{(1\,atm)}{\mathop{(pt){{H}_{2}}}}\,\left| \underset{(0.01M)}{\mathop{{{H}^{+}}}}\, \right|\left| \underset{0.1\,M}{\mathop{C{{u}^{2+}}}}\, \right|Cu $ $ (E_{C{{u}^{2+}}|Cu}^{o}=0.337\,V) $
The cell potential of the following cell at $ \text{25}{{\,}^{\text{o}}}\text{C} $ (in volts) is $ \underset{(1\,atm)}{\mathop{(pt){{H}_{2}}}}\,\left| \underset{(0.01M)}{\mathop{{{H}^{+}}}}\, \right|\left| \underset{0.1\,M}{\mathop{C{{u}^{2+}}}}\, \right|Cu $ $ (E_{C{{u}^{2+}}|Cu}^{o}=0.337\,V) $
Updated On: May 22, 2024
- $ 0.308 $
- $ 0.427 $
- $ -0.308 $
- $ 0.337 $
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The Correct Option is B
Solution and Explanation
$ \underset{(1\,atm)}{\mathop{(Pt){{H}_{2}}}}\,\left| {{\underset{(0.01\,M)}{\mathop{H}}\,}^{+}} \right|\left| \underset{(0.1\,M)}{\mathop{C{{u}^{2+}}}}\, \right|Cu $ For the above cell, the cell reaction is $ {{H}_{2}}+C{{u}^{2+}}\xrightarrow{{}}2{{H}^{+}}+Cu $ According to the Nemst equation $ {{E}_{cell}}=E_{C{{u}^{2+}}/Cu}^{o}-\frac{0.0591}{n}\log \frac{{{[{{H}^{+}}]}^{2}}}{[C{{u}^{2+}}]} $ $ =0.337-\frac{0.0591}{2}\log \frac{{{(0.01)}^{2}}}{(0.1)} $ $ =0.337-0.02955\log {{10}^{-3}} $ $ =0.337+0.08865 $ $ =0.42565\approx 0.427 $
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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.
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