Question

The EMF of the following cell is 0.22 volt.

Ag(s)|AgCl(s)|KCl(1M)|H+(1M)|H₂(g) (1atm); Pt(s)

Which of the following will decrease the EMF of cell.

• A. increasing pressure of H₂(g) from 1 atm to 2 atm
• B. increasing Cl⁻ concentration in Anodic compartment
• C. increasing H⁺ concentration in cathodic compartment
• D. Decreasing KCl concentration in Anodic compartment.

Answer 1

Answer: (A), (D)

Increasing pressure of H₂(g) from 1 atm to 2 atm OR Decreasing KCl concentration in Anodic compartment.

Answer 2

The given electrochemical cell is:

Ag(s)|AgCl(s)|KCl(1M)|H+(1M)|H₂(g) (1atm); Pt(s)

According to the IUPAC convention for cell notation, the anode (oxidation half-cell) is written on the left, and the cathode (reduction half-cell) is written on the right.

1. Anode (Oxidation): Ag electrode in contact with AgCl and Cl⁻ ions. Ag(s) + Cl⁻(aq) → AgCl(s) + e⁻

2. Cathode (Reduction): Standard Hydrogen Electrode (SHE). 2H⁺(aq) + 2e⁻ → H₂(g)

To get the overall cell reaction, we balance the electrons by multiplying the anode reaction by 2:

2Ag(s) + 2Cl⁻(aq) → 2AgCl(s) + 2e⁻ 2H⁺(aq) + 2e⁻ → H₂(g)

Overall Cell Reaction:

2Ag(s) + 2Cl⁻(aq) + 2H⁺(aq) → 2AgCl(s) + H₂(g)

Now, let's write the Nernst equation for this reaction. The number of electrons transferred (n) is 2. The reaction quotient Q is given by:

$Q = \frac{[AgCl(s)]^2 \cdot P_{H_2(g)}}{[Ag(s)]^2 \cdot [Cl^-(aq)]^2 \cdot [H^+(aq)]^2}$

Since Ag(s) and AgCl(s) are solids, their activities are considered as 1.

So, $Q = \frac{P_{H_2(g)}}{[Cl^-(aq)]^2 \cdot [H^+(aq)]^2}$

The Nernst equation is:

$E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q$ $E_{cell} = E^\circ_{cell} - \frac{RT}{2F} \ln \left( \frac{P_{H_2(g)}}{[Cl^-(aq)]^2 \cdot [H^+(aq)]^2} \right)$

Now let's analyze how changes in each option affect $E_{cell}$:

(A) increasing pressure of H₂(g) from 1 atm to 2 atm

If $P_{H_2(g)}$ increases, Q (which has $P_{H_2(g)}$ in the numerator) increases. According to the Nernst equation, an increase in Q makes the term $\frac{RT}{2F} \ln Q$ more positive. Since this term is subtracted from $E^\circ_{cell}$, the overall $E_{cell}$ will decrease.

(B) increasing Cl⁻ concentration in Anodic compartment

If $[Cl^-(aq)]$ increases, Q (which has $[Cl^-(aq)]$ in the denominator) decreases. A decrease in Q makes the term $\frac{RT}{2F} \ln Q$ less positive (or more negative). Since this term is subtracted from $E^\circ_{cell}$, the overall $E_{cell}$ will increase.

(C) increasing H⁺ concentration in cathodic compartment

If $[H^+(aq)]$ increases, Q (which has $[H^+(aq)]$ in the denominator) decreases. A decrease in Q makes the term $\frac{RT}{2F} \ln Q$ less positive (or more negative). Since this term is subtracted from $E^\circ_{cell}$, the overall $E_{cell}$ will increase.

(D) Decreasing KCl concentration in Anodic compartment.

KCl provides Cl⁻ ions. Decreasing KCl concentration means decreasing $[Cl^-(aq)]$. If $[Cl^-(aq)]$ decreases, Q (which has $[Cl^-(aq)]$ in the denominator) increases. An increase in Q makes the term $\frac{RT}{2F} \ln Q$ more positive. Since this term is subtracted from $E^\circ_{cell}$, the overall $E_{cell}$ will decrease.

Both options (A) and (D) lead to a decrease in the EMF of the cell.