For the given cell $\mathrm{Fe}^{2+}(\mathrm{eq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$ The standard cell potential of the above reaction is Given : $\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}$ $\mathrm{E}^{0}=\mathrm{xV}$ $\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}$ $\mathrm{E}^{0}=\mathrm{yV}$ $\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}$ $\mathrm{E}^{0}=\mathrm{zV}$

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Redox Reactions and Electrochemistry

For the given cell

Fe^{2 +} (eq) + Ag^{+} (aq) \to Fe^{3 +} (aq) + Ag (s)

The standard cell potential of the above reaction is
Given :

Ag^{+} + e^{-} \to Ag

E^{0} = xV

Fe^{2 +} + 2 e^{-} \to Fe

E^{0} = yV

Fe^{3 +} + 3 e^{-} \to Fe

E^{0} = zV

Answer: (B)

Electrode potentials cannot be added or subtracted directly across half-reactions with different numbers of electrons — their free energies (

Δ G^{\circ} = - n F E^{\circ}

) must be combined instead.

F e^{3 +} + 3 e^{-} \to F e, Δ G_{3}^{\circ} = - 3 F z

F e^{2 +} + 2 e^{-} \to F e, Δ G_{1}^{\circ} = - 2 F y

Subtracting the second from the first gives the desired one-electron couple,

F e^{3 +} + e^{-} \to F e^{2 +}

Δ G^{\circ} = Δ G_{3}^{\circ} - Δ G_{1}^{\circ} = - 3 F z - (- 2 F y) = 2 F y - 3 F z

Since this is a one-electron step,

Δ G^{\circ} = - F \cdot E_{F e^{3 +} / F e^{2 +}}^{\circ}

, so

E_{F e^{3 +} / F e^{2 +}}^{\circ} = 3 z - 2 y

For the overall cell reaction

F e^{2 +} + A g^{+} \to F e^{3 +} + A g

A g^{+}

is reduced at the cathode (

E^{\circ} = x

) while

F e^{2 +}

is oxidised to

F e^{3 +}

at the anode — the reverse of the couple just found:

E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = x - (3 z - 2 y) = x + 2 y - 3 z

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option