Question

For the reaction :
\(2AgCl(s) + H_2(g) (0.4 \text{ atm}) \rightarrow 2Ag(s) + 2H^+(0.1 \text{ M}) + 2Cl^-(0.2 \text{ M})\)
Calculate emf of the cell at \(25 \text{ } ^\circ\text{C}\).
Given : \(\Delta G^\circ = -43500 \text{ J mol}^{-1}\)
\([\log 10 = 1, 1 \text{ F} = 96500 \text{ C mol}^{-1}]\)

Hint:

When a gas is involved in a cell reaction, its activity is represented by its partial pressure in the reaction quotient expression. Don't forget to include the stoichiometric coefficients as powers in the \(Q\) expression.

Answer 1

Step 1: Understanding the Concept:
To find the non-standard EMF (\(E_{cell}\)), we first need the standard cell potential (\(E^\circ_{cell}\)), which can be derived from the standard Gibbs free energy change (\(\Delta G^\circ\)). Then, apply the Nernst equation incorporating the activities of gases (partial pressure) and aqueous ions.
Step 2: Detailed Explanation:
1. Calculate Standard EMF (\(E^\circ_{cell}\)):
\[ E^\circ_{cell} = \frac{-(-43500\text{ J/mol})}{2 \times 96500\text{ C/mol}} = \frac{43500}{193000} \approx 0.2254\text{ V} \]
2. Calculate Reaction Quotient (\(Q\)):
\[ Q = \frac{(0.1)^2 \times (0.2)^2}{0.4} = \frac{0.01 \times 0.04}{0.4} = \frac{0.0004}{0.4} = 0.001 = 10^{-3} \]
3. Calculate EMF (\(E_{cell}\)) using Nernst Equation:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{2} \log (10^{-3}) \]
\[ E_{cell} = 0.2254 - 0.0295 \times (-3 \log 10) \]
Since \(\log 10 = 1\):
\[ E_{cell} = 0.2254 + (0.0295 \times 3) = 0.2254 + 0.0885 = 0.3139\text{ V} \]
Step 3: Final Answer:
The EMF of the cell at \(25 \text{ } ^\circ\text{C}\) is \(0.3139\text{ V}\).