Calculate emf of the half cell given below: Pt(s) H₂(g, 2 atm) HCl (aq, 0.02 M)$$\text{Pt(s)} \mid H_2(\text{g, 2 atm}) \mid \text{HCl (aq, 0.02 M)}$$ E°H₂/H^+ = 0$$E^{\circ}_{H_2/H^+} = 0$$ (Given: (2.303 RT)/(F) = 0.059, 2 = 0.3010$\frac{2.303 RT}{F} = 0.059, \quad \log 2 = 0.3010$)

Related Formula

Core Logic

For Cu²⁺ + 2e^- arrow Cu(s)$Cu^{2+} + 2e^- \rightarrow Cu(s)$, n = 2$n = 2$. I = 1.5 A, t = 10 × 60 = 600 s$I = 1.5 \text{ A}, \quad t = 10 \times 60 = 600 \text{ s}$.

Step 1: Conclusion

Mass of copper deposited is 0.2938 g.

Pattern Recognition

Apply Faraday's First Law: w = Z I t = (M)/(n F) I t$w = Z I t = \frac{M}{n F} I t$.

Chapter Mix

Class 12 Chemistry: Electrochemistry