JEE Advanced 2025 — Chemistry Physical Chemistry

For the cell $\text{Zn}|\text{Zn}^{2+}(0.01\text{M})||\text{Cu}^{2+}(1\text{M})|\text{Cu}$, if $E^{\circ}_{\text{cell}} = 1.1\text{V}$, the EMF at $298\text{K}$ is:

$(R = 8.314 \text{ J mol}^{-1}\text{K}^{-1},\; F = 96500 \text{ C mol}^{-1})$