Given that $E°_{Ni^{2+}/Ni}=-0.25 V, E°_{Cu^{2+}/Cu} = +0.34 V$. The EMF of the cell containing nickel and copper electrodes in 1M solution is:

0.59 V

The correct answer is Option (2) → 0.59 V.

The electromotive force (EMF) of a cell can be calculated using the standard electrode potentials of the half-reactions. The cell setup involves nickel (Ni) and copper (Cu) electrodes, and the standard reduction potentials are given as:

\( E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.25 \, \text{V} \)

\( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \, \text{V} \)

The standard cell potential (\( E^\circ_{\text{cell}} \)) is calculated using the formula:

\(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\)

where:

Cathode is the electrode where reduction occurs (higher reduction potential),

Anode is the electrode where oxidation occurs (lower reduction potential).

Nickel half-cell reaction (anode): \( \text{Ni} \rightarrow \text{Ni}^{2+} + 2e^- \)

\( E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.25 \, \text{V} \)

Copper half-cell reaction (cathode): \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)

\( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \, \text{V} \)

Using the equation \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \):

\(E^\circ_{\text{cell}} = 0.34 \, \text{V} - (-0.25 \, \text{V})\)

\(E^\circ_{\text{cell}} = 0.34 \, \text{V} + 0.25 \, \text{V} = 0.59 \, \text{V}\)

Conclusion:

The EMF of the cell is 0.59 V.