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:
Answer & explanation
Correct answer: option 2
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.