Question

$\Delta$ G° of the cell reaction AgCl(s) + $\frac{1}{2}$H$_2$(g) $\rightarrow$ Ag(s) + H$^+$ +Cl$^-$ is -21.52 KJ $\Delta$G° of 2AgCl(s) +H$_2$(g) $\rightarrow$ 2Ag(s) +2H$^+$ +2Cl$^-$ is :

• A. -21.52 KJ
• B. -10.76 KJ
• C. -43.04 KJ
• D. 43.04 KJ

Answer 1

Answer: (C)

-43.04 KJ

Answer 2

The given reaction is: AgCl(s) + $\frac{1}{2}$H$_2$(g) $\rightarrow$ Ag(s) + H$^+$ +Cl$^-$

For this reaction, $\Delta$G° = -21.52 KJ.

The reaction for which $\Delta$G° is to be calculated is: 2AgCl(s) + H$_2$(g) $\rightarrow$ 2Ag(s) + 2H$^+$ + 2Cl$^-$

Upon comparing the two reactions, it is evident that the second reaction is exactly twice the first reaction. Since Gibbs free energy ($\Delta$G°) is an extensive property, its value is directly proportional to the stoichiometric coefficients of the reaction. If a reaction is multiplied by a factor 'n', its $\Delta$G° value is also multiplied by 'n'.

In this case, the second reaction is 2 times the first reaction. Therefore, $\Delta$G° for the second reaction = 2 $\times$ ($\Delta$G° for the first reaction)

$\Delta$G° = 2 $\times$ (-21.52 KJ)

$\Delta$G° = -43.04 KJ