E deg(Ce^4+/Ce^3+) = 1.61 V and E deg(Fe^3+/Fe^2+) = 0.77 V. A student writes a cell as Pt | Ce^3+ (1 M), Ce^4+ (1 M) || Fe^3+ (1 M), Fe^2+ (1 M) | Pt. Using the IUPAC convention E_cell = E_right - E_left, which statement is correct?

E_cell = -0.84 V; the spontaneous reaction is Ce^4+ + Fe^2+ -> Ce^3+ + Fe^3+

E_cell = +0.84 V; Fe^3+ oxidises Ce^3+

E_cell = -0.84 V; Fe^3+ oxidises Ce^3+

E_cell = +2.38 V; the cell is spontaneous as written

JEE Advanced levelSingle correct optionDifficulty 3 of 3ec036

$E^{\circ} (Ce X^{4 +} / Ce X^{3 +}) = 1.61 V$ and $E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = 0.77 V$ . A student writes a cell as $Pt ∣ Ce X^{3 +} (1 M), Ce X^{4 +} (1 M) ∣∣ Fe X^{3 +} (1 M), Fe X^{2 +} (1 M) ∣ Pt$ . Using the IUPAC convention $E_{cell} = E_{right} - E_{left}$ , which statement is correct?

A $E_{cell} = - 0.84 V$ ; the spontaneous reaction is $Ce X^{4 +} + Fe X^{2 +} Ce X^{3 +} + Fe X^{3 +}$
B $E_{cell} = + 0.84 V$ ; $Fe X^{3 +}$ oxidises $Ce X^{3 +}$
C $E_{cell} = - 0.84 V$ ; $Fe X^{3 +}$ oxidises $Ce X^{3 +}$
D $E_{cell} = + 2.38 V$ ; the cell is spontaneous as written

As written, the right electrode is

Fe X^{3 +} / Fe X^{2 +}

and the left is

Ce X^{4 +} / Ce X^{3 +}

, so

E_{cell} = 0.77 - 1.61 = - 0.84 V

. A negative EMF means the reaction implied by the notation (reduction on the right,

Fe X^{3 +} + Ce X^{3 +} Fe X^{2 +} + Ce X^{4 +}

) is non-spontaneous. The actual spontaneous reaction is the reverse: the stronger oxidant

Ce X^{4 +}

oxidises

Fe X^{2 +}

, with

E^{\circ} = + 0.84 V

. Potentials are never added (option D).