Read the passage carefully and answer the questions.

Electrochemistry is the study involving two types of processes. One type involves the production of electricity from the energy released during a spontaneous chemical reaction. Galvanic cells like Daniel cells are used to produce electricity from the redox reaction. Nernst equation of a galvanic cell gives the relation between the standard cell potential and the concentration of the species involved in the redox reaction. The second type of study involves the cells in which electrical energy is used to bring about the chemical reactions in a non-spontaneous process. The conductance of electrolytic solution depends on many factors including the nature of the electrolyte and the dimensions of the electrolytic cell.

The following equation represents the reaction taking place in a Galvanic cell:

$Mg(s) + 2Ag^+ (0.001M) → Mg^{2+}(0.01M) + 2Ag (s)$

Calculate its $E_{cell}$ at 298 K if $E^0_{cell} = 2.91 V$

2.79 V

The correct answer is Option (1) → 2.79 V

Given: $\text{E}_{\text{cell}}^{\circ}=2.91\,\text{V}$

Overall Reaction: $\text{Mg(s)}+2\text{Ag}^{+}(0.001\,\text{M})\rightarrow\text{Mg}^{2+}(0.01\,\text{M})+2\text{Ag(s)}$

Number of electrons transferred: $n=2 $

Reaction Quotient: $Q=\frac{[\text{Mg}^{2+}]}{[\text{Ag}^{+}]^2}=\frac{0.01}{(0.001)^2}=\frac{10^{-2}}{10^{-6}}=10^4$

Nernst Equation: $E_{\text{cell}}=E_{\text{cell}}^{\circ}-\frac{0.0592}{n}\log Q$

$E_{\text{cell}}=2.91-\frac{0.0592}{2}\log(10^4)$

$E_{\text{cell}}=2.91-(0.0296)(4)$

$E_{\text{cell}}=2.91-0.1184=2.7916\,\text{V}\approx2.79\,\text{V}$