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: Write hydrogen electrode reaction.
For hydrogen electrode: \[ \ce{2H+ + 2e^- -> H2(g)} \]

Step 2: Write Nernst equation.
For this reaction: \[ E=E^\circ-\frac{0.059}{2}\log\frac{P_{H_2}}{[H^+]^2} \] Since: \[ E^\circ=0 \] \[ E=-\frac{0.059}{2}\log\frac{P_{H_2}}{[H^+]^2} \]

Step 3: Substitute given values.
Given: \[ P_{H_2}=2 \] \[ [H^+]=0.02 \] \[ E=-\frac{0.059}{2}\log\frac{2}{(0.02)^2} \] \[ (0.02)^2=0.0004 \] \[ \frac{2}{0.0004}=5000 \] \[ E=-0.0295\log(5000) \]

Step 4: Calculate logarithm.
\[ 5000=5\times10^3 \] \[ \log5000=\log5+3 \] \[ \log5=\log\frac{10}{2}=1-\log2 \] \[ =1-0.3010=0.6990 \] So: \[ \log5000=0.6990+3=3.6990 \]

Step 5: Calculate emf.
\[ E=-0.0295\times3.6990 \] \[ E=-0.109V \] Therefore, the emf is: \[ -0.109V. \]