Question

For the reaction:
2 AgCl(s) + H₂(g) (0.4 atm) → 2 Ag(s) + 2 H⁺(0.1 M) + 2 Cl^-(0.2 M)
Calculate the emf of the cell at 25 °C.
Given: ΔG° = -43500 J mol^-1. [log 10 = 1, 1 F = 96500 C mol^-1]

Hint:

First get the standard cell potential from the standard free energy using ΔG° = -nFE° cell . Then apply the Nernst equation with the given concentrations and the hydrogen gas pressure to get the actual emf.

Answer 1

Step 1: Find the standard cell potential.
Use \(\Delta G^\circ = -nFE^\circ_{cell}\). Two electrons are transferred, so \(n = 2\).
\[E^\circ_{cell} = \dfrac{-\Delta G^\circ}{nF} = \dfrac{-(-43500)}{2 \times 96500} = \dfrac{43500}{193000} = 0.225\,V\]

Step 2: Write the reaction quotient.
For \(2AgCl + H_2 \rightarrow 2Ag + 2H^+ + 2Cl^-\):
\[Q = \dfrac{[H^+]^2[Cl^-]^2}{p_{H_2}}\]

Step 3: Put in the values.
\[Q = \dfrac{(0.1)^2 (0.2)^2}{0.4} = \dfrac{(0.01)(0.04)}{0.4} = \dfrac{0.0004}{0.4} = 10^{-3}\]

Step 4: Apply the Nernst equation.
\[E_{cell} = E^\circ_{cell} - \dfrac{0.059}{2}\log Q = 0.225 - \dfrac{0.059}{2}(-3)\]
\[E_{cell} = 0.225 + 0.0885 = 0.3135\,V\]

Step 5: State the result.
The actual emf is a little higher than the standard value because \(Q\) is less than 1.

Answer: \[\boxed{E_{cell} \approx 0.31\,V}\]