The standard Gibbs energy of the following reaction is \( -610\, kJ\, mol^{-1} \)
\( Mg(s) + 2Ag^+ (aq) → Mg^{2+}(aq) + 2Ag(s) \)
What is the standard EMF for the cell corresponding to the above reaction?
Given: Faraday's constant = \( 96500\, C\, mol^{-1} \)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → +3.16 V
The electrical work a galvanic cell can do equals the decrease in Gibbs energy of the cell reaction. Under standard conditions:
\( \displaystyle \Delta_r G^\circ = -nFE^\circ_{cell} \)
where n is the number of electrons transferred in the reaction as written. The two half-reactions are:
\( \displaystyle \text{Anode: } \ce{Mg(s) -> Mg^2+(aq) + 2e^-} \)
\( \displaystyle \text{Cathode: } \ce{2Ag^+(aq) + 2e^- -> 2Ag(s)} \)
Two electrons pass from Mg to the two \( \ce{Ag+} \) ions, so n = 2. Rearranging and substituting (\( \Delta_r G^\circ = -610\ \mathrm{kJ\,mol^{-1}} = -610000\ \mathrm{J\,mol^{-1}} \)):
\( \displaystyle E^\circ_{cell} = -\frac{\Delta_r G^\circ}{nF} \)
\( \displaystyle E^\circ_{cell} = -\frac{-610000\ \mathrm{J\,mol^{-1}}}{2 \times 96500\ \mathrm{C\,mol^{-1}}} \)
\( \displaystyle E^\circ_{cell} = \frac{610000}{193000}\ \mathrm{V} = +3.16\ \mathrm{V} \)
A negative \( \Delta_r G^\circ \) (a spontaneous reaction) always goes with a positive \( E^\circ_{cell} \). So the standard EMF of the cell is +3.16 V.