Question

The emf \( E^{\circ} \) of the following cells are:
\( \mathrm{Ag} \mid \mathrm{Ag}^{+} (1\text{ M}) \parallel \mathrm{Cu}^{2+} (1\text{ M}) \mid \mathrm{Cu}; \, E^{\circ} = -0.46 \text{ V} \)
\( \mathrm{Zn} \mid \mathrm{Zn}^{2+} (1\text{ M}) \parallel \mathrm{Cu}^{2+} (1\text{ M}) \mid \mathrm{Cu}; \, E^{\circ} = 1.10 \text{ V} \)
emf of the cell \( \mathrm{Zn} \mid \mathrm{Zn}^{2+} (1\text{ M}) \parallel \mathrm{Ag}^{+} (1\text{ M}) \mid \mathrm{Ag} \) is:

• A. $0.64 V$
• B. $1.10 V$
• C. $1.56 V$
• D. $-0.64 V$

Hint:

Target EMF = $EMF_{2} - EMF_{1}$ when cathodes are common.

Answer 1

The Correct Option is C

Step 1: Concept
Cell potential is the difference between reduction potentials: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Step 2: Analysis
From given data:
(i) $E^{\circ}_{Cu^{2+}/Cu} - E^{\circ}_{Ag^{+}/Ag} = -0.46V$.
(ii) $E^{\circ}_{Cu^{2+}/Cu} - E^{\circ}_{Zn^{2+}/Zn} = 1.10V$.
Subtracting (i) from (ii) gives: $E^{\circ}_{Ag^{+}/Ag} - E^{\circ}_{Zn^{2+}/Zn} = 1.10 - (-0.46)$.
Step 3: Conclusion
The resulting emf is $+1.56V$.
Final Answer: (C)