Calculate the standard cell potential (in V ) of the cell in which following reaction takes place: $\mathrm{Fe}^{2}+(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$ Given that: $\mathrm{E}^{0}_{{}_{{\mathrm{Ag}^+} / \mathrm{Ag}}}=\mathrm{xV}$ $\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{0}=\mathrm{yV}$ $\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}}^{0}=\mathrm{zV}$

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Redox Reactions and Electrochemistry

Calculate the standard cell potential (in V ) of the cell in which following reaction takes place:

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

Given that:

E_{_{Ag^{+} / Ag}}^{0} = xV

E_{Fe^{2 +} / Fe}^{0} = yV

E_{Fe^{3 +} / Fe}^{0} = zV

Answer: (D)

It is cleanest to work through free energies, since

Δ G

is additive while

E^{\circ}

is not, using

Δ G^{\circ} = - n F E^{\circ}

. For

Fe X^{2 +} + 2 e X^{-} Fe

Δ G_{1}^{\circ} = - 2 F y

; for

Fe X^{3 +} + 3 e X^{-} Fe

Δ G_{2}^{\circ} = - 3 F z

. Subtracting gives

Fe X^{2 +} Fe X^{3 +} + e X^{-}

with

Δ G^{\circ} = - 2 F y + 3 F z

. The silver half-reaction

Ag X^{+} + e X^{-} Ag

has

Δ G_{3}^{\circ} = - F x

. Adding the iron oxidation and the silver reduction for the overall cell,

Δ G_{cell}^{\circ} = - 2 F y + 3 F z - F x = - F E_{cell}^{\circ}

. Dividing by

- F

E_{cell}^{\circ} = x + 2 y - 3 z

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option