Question

Calculate emf and \(\Delta G\) for the following cell at \(298 \text{ K}\) :
\(Mg(s) / Mg^{2+}(0.01 \text{ M}) // Ag^+(0.001 \text{ M}) / Ag(s)\)
Given : \(E^\circ_{Mg^{2+}/Mg} = -2.37 \text{ V}\), \(E^\circ_{Ag^+/Ag} = +0.80 \text{ V}\)
\([1 \text{ F} = 96500 \text{ C mol}^{-1}, \log 10 = 1]\)

Hint:

Always ensure the stoichiometry is balanced to get the correct exponent for concentration terms in the log quotient. Squaring \([Ag^+]\) is crucial here because of the \(2Ag^+\) in the balanced equation.

Answer 1

Step 1: Understanding the Concept:
For a galvanic cell with non-standard concentrations, the electromotive force (EMF) is calculated using the Nernst equation. The Gibbs free energy change (\(\Delta G\)) is then calculated using the relationship \(\Delta G = -nFE_{cell}\).
Step 2: Detailed Explanation:
1. Calculate Standard EMF (\(E^\circ_{cell}\)):
\[ E^\circ_{cell} = 0.80\text{ V} - (-2.37\text{ V}) = 3.17\text{ V} \]
2. Calculate EMF (\(E_{cell}\)) using Nernst Equation:
\[ E_{cell} = 3.17 - \frac{0.059}{2} \log \frac{0.01}{(0.001)^2} \]
\[ E_{cell} = 3.17 - 0.0295 \log \frac{10^{-2}}{(10^{-3})^2} \]
\[ E_{cell} = 3.17 - 0.0295 \log \frac{10^{-2}}{10^{-6}} \]
\[ E_{cell} = 3.17 - 0.0295 \log 10^4 \]
Since \(\log 10^4 = 4 \log 10 = 4\):
\[ E_{cell} = 3.17 - 0.0295 \times 4 = 3.17 - 0.118 = 3.052\text{ V} \]
3. Calculate Gibbs Free Energy (\(\Delta G\)):
\[ \Delta G = -nFE_{cell} = -2 \times 96500 \times 3.052 \]
\[ \Delta G = -193000 \times 3.052 = -589036\text{ J mol}^{-1} \]
To convert to kJ:
\[ \Delta G = -589.036\text{ kJ mol}^{-1} \]
Step 3: Final Answer:
The EMF of the cell is \(3.052\text{ V}\) and the Gibbs free energy change \(\Delta G\) is \(-589.036\text{ kJ mol}^{-1}\).