What is the emf of the cell of the given reaction?
\(Mg(s) | Mg^{2+} (0.1 M) ||Ag^+ (1 × 10^{−4} M) | Ag\)
\(E^0_{Ag^+/ Ag} = 0.8V\), \(E^0_{Mg^{2+}/ Mg} = −2.37 V\)
Answer & explanation
Correct answer: option 1
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, \(E_{cell} = 2.96 V\)