Q55 (2024)
Calculate emf of the half cell given below:
$$\text{Pt(s)} \mid H_2(\text{g, 2 atm}) \mid \text{HCl (aq, 0.02 M)}$$
$$E^{\circ}_{H_2/H^+} = 0$$
(Given: $\frac{2.303 RT}{F} = 0.059, \quad \log 2 = 0.3010$)
- (1) 0.109 V
- (2) 0.035 V
- (3) -0.035 V
- (4) -0.109 V
### Related Formula
$$H_2(g) \rightarrow 2H^+(aq) + 2e^-$$
$$E = E^{\circ} - \frac{0.059}{n} \log \frac{[H^+]^2}{P_{H_2}}$$
### Core Logic
$$[H^+] = 0.02 \text{ M}, \quad P_{H_2} = 2 \text{ atm}, \quad n = 2$$
$$E = 0 - \frac{0.059}{2} \log \frac{(0.02)^2}{2} = -0.0295 \log (2 \times 10^{-4})$$
$$E = -0.0295 (\log 2 - 4) = -0.0295 (0.3010 - 4) = -0.0295 (-3.699) = +0.109 \text{ V}$$
### Step 1: Final Value
Half cell EMF is $+0.109 \text{ V}$.
### Pattern Recognition
For oxidation half cell $H_2 \rightarrow 2H^+ + 2e^-$, Nernst equation uses ratio $[H^+]^2 / P_{H_2}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
