Question

The zinc/silver oxide cell is used in electric watches. The reaction is as follows:
$Z{n^{2 + }} + 2{e^ - } \to Zn;{E^0} =$ -0.76 V
$A{g_2}O + {H_2}O \to 2Ag + O{H^ - };{E^0} =$ 0.344 V
If F is 96500 $mo{l^{ - 1}}$ , then $\Delta {G^0}$ of the cell is:
A.413 kJ/mol
B.113 kJ/mol
C.213 kJ/mol
D.313 kJ/mol

Answer 1

Nernst equation can be basically understood as an equation that helps us relate the reduction potential of an electrochemical cell to the standard electrode potential as well as the temperature of the system.

Complete step by step answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
Gibbs free energy can be understood as the energy which is associated with a given chemical reaction that can be used to do work. For a given system, the free energy can be calculated as the sum of the enthalpy, and the product of the temperature and entropy of the system. Under standard conditions, Nernst equation can be derived from Gibbs free energy. This Nernst equation can be mathematically represented as:
${E^0} = E_{reduction}^0 + E_{oxidation}^0$
Gibbs energy can be related to the Enthalpy of the reaction under general conditions via
$\Delta G = - nFE$ ;
Where ‘n’ is the number of electrons transferred in the chemical reaction, ‘F’ is the Faraday constant, E if the potential difference. This equation under standard conditions can be written as:
$\Delta {G^0} = - nF{E^0}$
In the given question,
$Z{n^{2 + }} + 2{e^ - } \to Zn;{E^0} =$ -0.76 V
$A{g_2}O + {H_2}O \to 2Ag + O{H^ - };{E^0} =$ 0.344 V
Hence,
${E^0} = E_{reduction}^0 + E_{oxidation}^0$
${E^0} = 0.344-\left( { - 0.76} \right) = 0.344 + 0.76 =$ 1.104 V
Substituting this value in the other equation, we get
$\Delta {G^0} = - nF{E^0}$
$\Delta {G^0}\; = - \left( 2 \right)\left( {96500} \right)\left( {1.104} \right)$
$\Delta {G^0}$ = 213 kJ/mol

Hence, Option C is the correct option.

Note:
The sign preceding the value of the Gibbs energy determines the nature of the given reaction. If this value is negative, the reaction is spontaneous, and if this value is positive, then the reaction is non - spontaneous.