Calculate the EMF of the cell: Zn|Zn²⁺(0.1 M)||Cu²⁺(0.01 M)|Cu Given: E°Zn²⁺/Zn = -0.76 V, E°Cu²⁺/Cu = +0.34 V — JEE Chemistry

$$E°_{cell} = E°_{cathode} - E°_{anode} = 0.34 - (-0.76) = 1.10 V$$

Using Nernst equation ($n = 2$, $T = 298$ K):

$$E = E° - \frac{0.0591}{n}\log Q = 1.10 - \frac{0.0591}{2}\log\frac{[Zn^{2+}]}{[Cu^{2+}]}$$

$$= 1.10 - 0.02955\times\log\frac{0.1}{0.01} = 1.10 - 0.02955\times1$$

$$E = 1.07 V$$