Question

Calculate emf and ΔG for the following cell at 298 K:
Mg(s) | Mg²⁺(0.01 M) || Ag⁺(0.001 M) | Ag(s)
Given: E°_Mg²⁺/Mg = -2.37 V, E°_Ag⁺/Ag = +0.80 V. [1 F = 96500 C mol^-1, log 10 = 1]

Hint:

The cell emf is found by first getting the standard cell potential E° cell = E° cathode - E° anode , then correcting for the actual concentrations using the Nernst equation.

Answer 1

Step 1: Find the standard cell potential.
Magnesium has the more negative potential, so it is the anode; silver is the cathode.
\[E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-2.37) = 3.17\,V\]

Step 2: Write the cell reaction and find n.
\[Mg + 2Ag^+ \rightarrow Mg^{2+} + 2Ag\]
Two electrons are transferred, so \(n = 2\).

Step 3: Set up the reaction quotient.
\[Q = \dfrac{[Mg^{2+}]}{[Ag^+]^2} = \dfrac{0.01}{(0.001)^2} = \dfrac{0.01}{10^{-6}} = 10^{4}\]

Step 4: Apply the Nernst equation.
\[E_{cell} = E^\circ_{cell} - \dfrac{0.059}{n}\log Q = 3.17 - \dfrac{0.059}{2}(4)\]
\[E_{cell} = 3.17 - 0.118 = 3.052\,V\]

Step 5: Find the free energy change.
\[\Delta G = -nFE_{cell} = -(2)(96500)(3.052)\]
\[\Delta G = -589036\,J \approx -589\,kJ\]

Answer: \[\boxed{E_{cell} \approx 3.05\,V, \quad \Delta G \approx -589\,kJ\,mol^{-1}}\]