What is the emf of the following cell at 298 K?Mg(s) | Mg2+ (0.1 M) |…

The correct answer is Option (1) → 2.96 V

The emf of a cell depends on the concentrations of the ions. The Nernst equation gives it from the standard emf, the number of electrons transferred (n) and the reaction quotient. At 298 K:

    \( \displaystyle E_{\text{cell}}=E^{\circ}_{\text{cell}}-\frac{0.059\,\mathrm{V}}{n}\log\frac{[\text{products}]}{[\text{reactants}]} \)

In the cell notation the left electrode is the anode (oxidation) and the right electrode is the cathode (reduction):

    Anode: \( \displaystyle \ce{Mg(s) -> Mg^2+(aq) + 2e^-} \)
    Cathode: \( \displaystyle \ce{2Ag^+(aq) + 2e^- -> 2Ag(s)} \)
    Cell reaction: \( \displaystyle \ce{Mg(s) + 2Ag^+(aq) -> Mg^2+(aq) + 2Ag(s)} \)

Two electrons are transferred, so n = 2. Each \( \ce{Ag^+} \) takes one electron, so the silver half-reaction is doubled and \( [\ce{Ag^+}] \) is squared in the reaction quotient. Solids are left out of it.

The standard emf is the cathode potential minus the anode potential:

    \( \displaystyle E^{\circ}_{\text{cell}}=E^{\circ}_{\ce{Ag^+/Ag}}-E^{\circ}_{\ce{Mg^2+/Mg}} \)
    \( \displaystyle E^{\circ}_{\text{cell}}=0.80-(-2.37)=3.17\,\mathrm{V} \)

Now the Nernst equation for this cell:

    \( \displaystyle E_{\text{cell}}=E^{\circ}_{\text{cell}}-\frac{0.059}{2}\log\frac{[\ce{Mg^2+}]}{[\ce{Ag^+}]^2} \)

    \( \displaystyle \frac{[\ce{Mg^2+}]}{[\ce{Ag^+}]^2}=\frac{0.1}{(1\times10^{-4})^2}=\frac{10^{-1}}{10^{-8}}=10^{7} \)

    \( \displaystyle E_{\text{cell}}=3.17-\frac{0.059}{2}\times\log 10^{7} \)

    \( \displaystyle E_{\text{cell}}=3.17-0.0295\times7=3.17-0.21 \)
    \( \displaystyle E_{\text{cell}}=2.96\,\mathrm{V} \)

The emf is a little below \( E^{\circ}_{\text{cell}} \) because the reactant \( \ce{Ag^+} \) is very dilute, so the reaction quotient is greater than 1.