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

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: \(E^\circ_{Mg^{2+}/Mg} = -2.37 \text{ V}\), \(E^\circ_{Ag^+/Ag} = +0.80 \text{ V}\)
\([1 \text{ F} = 96500 \text{ C mol}^{-1}, \log 10 = 1]\)
Solution:
The cell reaction is: \[ \text{Mg}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Mg}^{2+}(aq) + 2\text{Ag}(s) \] The standard cell potential (\(E^\circ_{\text{cell}}\)) is given by:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] Here, the anode is the magnesium electrode and the cathode is the silver electrode. So:
\[ E^\circ_{\text{cell}} = E^\circ_{Ag^+/Ag} - E^\circ_{Mg^{2+}/Mg} \] Substituting the values:
\[ E^\circ_{\text{cell}} = 0.80 \text{ V} - (-2.37 \text{ V}) = 3.17 \text{ V} \] Now, we calculate the actual cell potential (\(E_{\text{cell}}\)) using the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \] Where:
- \(n = 2\) (because 2 electrons are transferred in the reaction),
- \(Q = \frac{[\text{Mg}^{2+}]}{[\text{Ag}^+]^2}\),
- \([\text{Mg}^{2+}] = 0.01 \, \text{M}, [\text{Ag}^+] = 0.001 \, \text{M}\).
So:
\[ Q = \frac{0.01}{(0.001)^2} = 10000 \] Now, substituting the values into the Nernst equation:
\[ E_{\text{cell}} = 3.17 \text{ V} - \frac{0.0591}{2} \log 10000 \] \[ \log 10000 = 4 \] \[ E_{\text{cell}} = 3.17 \text{ V} - \frac{0.0591}{2} \times 4 \] \[ E_{\text{cell}} = 3.17 \text{ V} - 0.1182 \text{ V} = 3.0518 \text{ V} \] So, the cell potential is:
\[ E_{\text{cell}} \approx 3.05 \text{ V} \] Now, calculate \(\Delta G\):
The change in Gibbs free energy is related to the cell potential by the equation:
\[ \Delta G = -nFE_{\text{cell}} \] Substituting the values:
\[ \Delta G = -2 \times 96500 \, \text{C/mol} \times 3.05 \, \text{V} \] \[ \Delta G = -2 \times 96500 \times 3.05 = -589,030 \, \text{J/mol} \] \[ \Delta G \approx -589 \, \text{kJ/mol} \] Final Answer:
The emf of the cell is approximately \( 3.05 \, \text{V} \) and the change in Gibbs free energy, \(\Delta G\), is approximately \( -589 \, \text{kJ/mol} \).