Question:
What is the electrode potential (in $ V $ ) of the following electrode at $ {{25}^{o}}C $ ? $ N{{i}^{2+}}(0.1\,\,M)|Ni(s) $ (Standard reaction potential of $ N{{i}^{2+}}|Ni $ is $ -0.25\,\,V,\,\,\frac{2.303RT}{F}=0.06) $
What is the electrode potential (in $ V $ ) of the following electrode at $ {{25}^{o}}C $ ? $ N{{i}^{2+}}(0.1\,\,M)|Ni(s) $ (Standard reaction potential of $ N{{i}^{2+}}|Ni $ is $ -0.25\,\,V,\,\,\frac{2.303RT}{F}=0.06) $
Updated On: Jul 28, 2022
- $ -0.28\,\,V $
- $ -0.34\,\,V $
- $ -0.82\,\,V $
- $ -0.22\,\,V $
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The Correct Option is A
Solution and Explanation
$ E_{cell}^{o}=E_{cell}^{o}-\frac{2.303RT}{nF}\log \left[ \frac{[Ni]}{[N{{i}^{2+}}]} \right] $ $ =-0.25-\frac{0.06}{2}\log \left( \frac{1}{0.1} \right) $ $ (\because \,\,n=2) $ $ =-0.25-0.03\times 1 $ $ {{E}_{cell}}=-0.28\,\,V $
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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.
