Question

The Following standard free energy change for the reaction
${H_2}(g) + 2AgCl(s) \to 1Ag(s) + 2{H^ + }(aq.) + 2C{l^ - }(aq.)$
is$- 10.26kcalmo{l^{ - 1}}$ at ${25^ \circ }C$. A cell using the above reaction is operated at ${25^ \circ }C$ under
${P_{H2}} = 1atm.[{H^ + }]$and$[C{l^{ - ]}}] = 0.1$. Calculate the emf of the cell.
A. $0.140volt$
b. $0.240volt$
C. $0.340volt$
D. $0.440volt$

Answer 1

Hint : In order to solve this question we must know about the concept of Gibbs free energy. It’s said the max. amount of work$({w_{\max }} = \Delta {G^0})$is equal to the product of cell potential (${E^0}$) and the total amount of transferred charge during the reaction $\left( {nF} \right)$→$\Delta {G^ \circ } = - nF{E^ \circ }$.

Complete step-by-step solution:
From this question, here we get,
$\Delta {G^0} = - 10.26Kcalmo{l^{ - 1}}$
And From the equation of Gibbs free energy, we also know, $\Delta {G^0} = - nF{E^0}$
Now we can write,

$${E^0} = - \frac{{10.26 \times {{10}^3} \times 4.159}}{{ - 2 \times 96500}} \\\ \Rightarrow {E^0} = 0.22109 \\\$$

Therefore, applying the Nernst equation we get,

$$E = {E^0} - \frac{{0.0592}}{n}\log \frac{{{{({H^ + })}^2} * {{(C{l^ - })}^2}}}{{{P_{H2}}}} \\\ \Rightarrow E = 0.22109 + 0.1182 \\\ \Rightarrow E = 0.339 \\\ \therefore E \simeq 0.34volt \\\$$

So, option c) $0.340volt$ is correct.

Note: Unconstrained redox reaction has a$\left( { - ve} \right)\Delta {G^0}$and become a$\left( { + ve} \right){E_{cel}}_l$. Huge equilibrium constants relate to the enormous $\left( { + ve} \right)$value of${E^0}$.