The unit of E_cell is

The correct answer is Option (3) → \(V\)

The cell potential (emf) is the difference between the electrode potentials of the cathode and the anode:

    \( \displaystyle E_{cell} = E_{right} - E_{left} = E_{cathode} - E_{anode} \)

A potential difference is the work done per unit charge, joule per coulomb, which is the volt (V). For example, \( E_{cell} = 1.1\ \mathrm{V} \) for the Daniell cell. It is also linked to Gibbs energy by \( \Delta_r G = -nFE_{cell} \), where J mol⁻¹ divided by C mol⁻¹ again gives J C⁻¹ = V.

Why the other options are incorrect

• Option (1) \(V\, m^{-1}\) — Incorrect. This is the unit of electric field strength (potential per unit length), not of a potential.

• Option (2) \(S\, cm^{-1}\) — Incorrect. This is the unit of conductivity, κ.

• Option (4) \(S\, cm^{-2}\,mol^{-1}\) — Incorrect. This is not a standard unit; molar conductivity is S cm² mol⁻¹.

NCERT (2026-27 edition):
Galvanic Cells – “… The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts. The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the cell electromotive force (emf) of the cell when no current is drawn through the cell.”