What is the emf of the cell of the given reaction?

\(Mg(s) | Mg^{2+} (0.1 M) ||Ag^+ (1 × 10^{−4} M) | Ag\)

\(E^0_{Ag^+/ Ag} = 0.8V\), \(E^0_{Mg^{2+}/ Mg} = −2.37 V\)

2.96V

The correct answer is option 1. 2.96V.

\(E^0_{cell} = E^0_{cathode} − E^0_{anode}\)

or, \(E^0_{cell} = 0.80 − (− 2.37) = 3.17 V\)

The cell reaction will be,

\(Mg  +  2Ag^+  \rightarrow  2Ag  +  Mg^{2+}\)

From the Nernst equation, we know,

\(E_{cell} = E^0_{cell} − \frac{0.0591}{2}log\frac{[Mg^{2+}]}{[Ag^+]^2}\)

or, \(E_{cell} = 3.17 − \frac{0.0591}{2}log\frac{10^{−1}}{(10^{−4})^2}\)

or, \(E_{cell} = 3.17 − \frac{0.0591}{2} ×7\)

or, \(E_{cell} = 3.17 − 0.20685\)

or, \(E_{cell} = 2.96 V\)