Consider the following $\mathrm{E}^{\circ}$ values $\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.77 \mathrm{~V}$ $\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V}$ Under standard conditions the potential for the reaction $\mathrm{Sn}(\mathrm{s})+2 \mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})$ is

JEE Main · 2004 · 24 Apr 2004 · Offline · Redox Reactions and Electrochemistry

Consider the following

E^{\circ}

values

E_{Fe^{3 +} / Fe^{2 +}}^{\circ} = 0.77 V

E_{Sn^{2 +} / Sn}^{\circ} = - 0.14 V

Under standard conditions the potential for the reaction

Sn (s) + 2 Fe^{3 +} (aq) ⟶ 2 Fe^{2 +} (aq) + Sn^{2 +} (aq)

Answer: (C)

Combine the half-cells with

E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}

. Iron(III) is reduced to iron(II) at the cathode with

E^{\circ} = 0.77 V

, while tin metal is oxidised to

Sn X^{2 +}

at the anode with

E^{\circ} = - 0.14 V

. Subtracting,

E_{ce l l}^{\circ} = 0.77 - (- 0.14) = 0.91 V

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option