Given that \(E^\circ _{Ni^{2+}/Ni}=-0.25 V, E^\circ _{Cu^{2+}/Cu} = +0.34 V\). 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 electrode with the higher (more positive) reduction potential is the cathode. Here copper (+0.34 V) is the cathode and nickel (−0.25 V) the anode:
\( \displaystyle \text{Anode: }\ce{Ni(s) -> Ni^2+(aq) + 2e^-} \)
\( \displaystyle \text{Cathode: }\ce{Cu^2+(aq) + 2e^- -> Cu(s)} \)
\( \displaystyle \text{Cell: }\ce{Ni(s) + Cu^2+(aq) -> Ni^2+(aq) + Cu(s)} \)
\( \displaystyle E^\circ_{\text{cell}} = E^\circ_{\text{right}} - E^\circ_{\text{left}} = (+0.34) - (-0.25) = 0.59\ \mathrm{V} \)
With 1 M solutions the EMF equals the standard EMF, 0.59 V.
NCERT (2026-27 edition):
Galvanic Cells – “… A galvanic cell is generally represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by a salt bridge. Under this convention the emf of the cell is positive and is given by the potential of the half-cell on the right hand side minus the potential of the half-cell on the left hand side i.e., Ecell = Eright – Eleft”