Question:medium

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:

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Target EMF = $EMF_{2} - EMF_{1}$ when cathodes are common.
Updated On: Apr 8, 2026
  • $0.64 V$
  • $1.10 V$
  • $1.56 V$
  • $-0.64 V$
Show Solution

The Correct Option is C

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