A galvanic cell is represented as Ni(s)|Ni2+(aq)||Ag+(aq)|Ag(s). Whic…

The correct answer is Option (2) → Nickel electrode is the anode, and E_cell = E_Ag⁺|Ag − E_Ni²⁺|Ni

By convention a galvanic cell is written with the anode on the left and the cathode on the right, and its emf is E_cell = E_right − E_left, the electrode potential of the cathode minus that of the anode.

• Left half-cell Ni(s)|Ni²⁺(aq): the anode, where nickel is oxidised:

      \( \ce{Ni(s) -> Ni^2+(aq) + 2e^-} \)

• Right half-cell Ag⁺(aq)|Ag(s): the cathode, where silver ions are reduced:

      \( \ce{2Ag+(aq) + 2e^- -> 2Ag(s)} \)

• Cell reaction:

      \( \ce{Ni(s) + 2Ag+(aq) -> Ni^2+(aq) + 2Ag(s)} \)

  E_cell = E_right − E_left = E_Ag⁺|Ag − E_Ni²⁺|Ni

Why the other options are incorrect

Option (1) Nickel electrode is the cathode, and E_cell = E_Ni²⁺|Ni − E_Ag⁺|Ag — Incorrect. Nickel, written on the left, is the anode, not the cathode, and the expression is anode minus cathode, which is the wrong way round.

Option (3) Nickel electrode is the anode, and E_cell = E_Ni²⁺|Ni − E_Ag⁺|Ag — Incorrect. The anode is right, but E_cell is E_right − E_left; subtracting the cathode from the anode gives the wrong sign.

Option (4) Nickel electrode is the cathode, and E_cell = E_Ag⁺|Ag − E_Ni²⁺|Ni — Incorrect. The expression is right, but the nickel electrode is the anode; the silver electrode is the cathode.

NCERT (2026-27 edition):
2.2 Galvanic Cells – “…It is now an accepted convention that we keep the anode on the left and the cathode on the right while representing the galvanic cell. … 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., E_cell = E_right – E_left”

2.3 Nernst Equation – “…The cell reaction is Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s)”