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\[{E^0}\].