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  • In a concentration cell, the concentration of  ions in one half-cel

In a concentration cell, the concentration of \mathrm{Ag}^{+} ions in one half-cell is 10^{-3}M and in the other half-cell is 10^{-6}M. What is the cell potential at 25^{\circ} \mathrm{C} ? \text { (Given: }\left.E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=0.80 \mathrm{~V} \text { at } 25^{\circ} \mathrm{C}\right)

Option: 1

0.225 V


Option: 2

0.75 V


Option: 3

0.80 V


Option: 4

0.85 V


Answers (1)

best_answer

The cell potential of a concentration cell is given by the Nernst equation, which is:

E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{R T}{n F} \ln Q

Where E_{\text {cell }}^{\circ} is the standard cell potential, R is the gas constant, T is the temperature in kelvin, n is the number of electrons transferred in the cell reaction, Fis the Faraday constant, and Q is the reaction quotient. For a concentration cell, the reaction quotient is the ratio of the concentrations of the same ion on the two electrodes:

Q=\frac{\left[\mathrm{Ag}^{+}\right]_{\text {high }}}{\left[\mathrm{Ag}^{+}\right]_{\text {low }}}

Substituting the given values in the Nernst equation, we get:

E_{\text {cell }}=0.80-\frac{0.025}{1} \ln \frac{10^{-3}}{10^{-6}}

Simplifying, we get:

E_{\text {cell }}=0.80-0.025 \ln 1000=0.80-0.575=0.225 \mathrm{~V}

Posted by

Suraj Bhandari

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