Question

Calculate emf of the half-cell given below: \[ Pt(s)\mid H_2(g,2\text{ atm})\mid HCl(aq,0.02M) \] \[ E^\circ_{H^+/H_2}=0V,\qquad \frac{2.303RT}{F}=0.059 \] \[ \log 2=0.3010 \]

• A. \(0.035V\)
• B. \(-0.035V\)
• C. \(-0.109V\)
• D. \(0.109V\)

Hint:

For hydrogen electrode, use \(E=E^\circ-\frac{0.059}{2}\log\frac{P_{H_2}}{[H^+]^2}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The goal is to find the electrode potential ($E$) of a hydrogen half-cell that is not at standard conditions. In a standard hydrogen electrode (SHE), the pressure is $1$ atm and the concentration of $H^+$ is $1 M$. However, here the pressure is $2$ atm and the $HCl$ concentration is $0.02 M$. We must apply the Nernst equation to find how these non-standard parameters affect the half-cell potential.
Step 2: Key Formula or Approach:
The reduction reaction for the hydrogen electrode is: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$. The Nernst Equation for a half-cell is: \[ E = E^\circ - \frac{2.303RT}{nF} \log \frac{P_{H_2}}{[H^+]^2} \]
$E^\circ = 0 V$ (by convention for SHE).
$n = 2$ (number of electrons transferred).
$0.059$ is given as the value for $\frac{2.303RT}{F}$.
Step 3: Detailed Explanation:

Parameters: $P_{H_2} = 2 \text{ atm}$ and $[H^+] = 0.02 M$ (as $HCl$ is a strong monoprotic acid).
Substitute into Formula: \[ E = 0 - \frac{0.059}{2} \log \frac{2}{(0.02)^2} \]
Simplify the Logarithmic Term: $(0.02)^2 = 0.0004 = 4 \times 10^{-4}$. So, $\frac{2}{4 \times 10^{-4}} = 0.5 \times 10^4 = 5000$.
Calculate Log Value: $\log(5000) = \log(5 \times 10^3) = \log 5 + 3$. Given $\log 2 = 0.3010$, we can find $\log 5$ because $\log 5 = \log(10/2) = \log 10 - \log 2 = 1 - 0.3010 = 0.6990$. Therefore, $\log(5000) = 0.6990 + 3 = 3.6990$.
Final Potential: \[ E = -0.0295 \times 3.6990 \approx -0.10912 V \]
Rounding gives $-0.109 V$.
Step 4: Final Answer:
The emf of the half-cell is $-0.109 V$.