Question

When two half cells of electrode potential of ${{E}_{1}}$ and ${{E}_{2}}$ are combined to form a half cell of electrode potential ${{E}_{3}}$, then
(When ${{n}_{1}}$,${{n}_{2}}$ and ${{n}_{3}}$ are no. of electrons exchanged in first, second and combined half – cells):
(a)${{E}_{3}}={{E}_{2}}-{{E}_{1}}$
(b) ${{E}_{3}}=\frac{{{E}_{1}}{{n}_{1}}+{{E}_{2}}{{n}_{2}}}{{{n}_{3}}}$
(c) ${{E}_{3}}=\frac{{{E}_{1}}{{n}_{1}}+{{E}_{2}}{{n}_{2}}}{{{n}_{3}}^{2}}$
(d) ${{E}_{3}}={{E}_{1}}+{{E}_{2}}$

Answer 1

Hint: Electrode potential is the potential difference set up between the electrodes in an electrochemical cell that is used to obtain electrical energy. Electrical work done can be equal to the electrical work done multiplied by total charge passed. Gibbs energy tells the useful work done by the electrochemical cell. Its value is $\Delta G=-nF{{E}_{cell}}$.

Complete answer:
Electrochemical cell is the cell that is used to convert the chemical energy into electrical energy. It involves electrodes dipped in a salt solution. The electrodes exert some potential difference due to the movement of electrons that create its cell potential denoted as E or ${{E}_{cell}}$.
The useful work done by an electrochemical cell can be calculated by Gibbs energy as$\Delta G=-nF{{E}_{cell}}$, where n is the number of electrons, F is Faraday’s constant and E is cell potential.
Now, if a combination of two cells is used, then the resultant work will be the addition of the work done by both cells. We have given two half cells of electrode potential of ${{E}_{1}}$ and ${{E}_{2}}$ that are combined to form a half cell of electrode potential${{E}_{3}}$. So, the resultant${{E}_{3}}$will be calculated by the sum of Gibbs energy of both cells as:
$\Delta {{G}_{3}}=\Delta {{G}_{1}}+\Delta {{G}_{2}}$ , substituting $\Delta G=-nF{{E}_{cell}}$
$-{{n}_{3}}F{{E}_{3}}=-{{n}_{1}}F{{E}_{1}}-{{n}_{2}}F{{E}_{2}}$, cancelling out the Faraday’s constant as common and then rearranging for ${{E}_{3}}$, we have
${{E}_{3}}=\frac{{{n}_{1}}{{E}_{1}}+{{n}_{2}}{{E}_{2}}}{{{n}_{3}}}$ where, ${{n}_{1}}$,${{n}_{2}}$ and${{n}_{3}}$are no. of electrons exchanged in first, second and combined half – cells.

Hence, the correct relationship is ${{E}_{3}}=\frac{{{E}_{1}}{{n}_{1}}+{{E}_{2}}{{n}_{2}}}{{{n}_{3}}}$, so option (b) is correct.

Note:
Electrode potential is an intensive property meaning it is independent upon the amount of electrons, while Gibbs energy is an extensive property and depends on the amount of electrons. If the concentration of all the species is taken as unity then at standard conditions, $\Delta {{G}^{o}}=-nF{{E}^{o}}_{cell}$. F is Faraday’s constant having value 96500 C.