Excess-free zinc is added to a solution containing 0.20 mol of Fe^3+ until all iron is deposited as the metal. Use E deg(Zn^2+/Zn) = -0.76 V, E deg(Fe^2+/Fe) = -0.44 V, E deg(Fe^3+/Fe^2+) = +0.77 V and Zn = 65. The mass of zinc consumed is

JEE Advanced levelSingle correct optionDifficulty 3 of 3df077

Excess-free zinc is added to a solution containing $0.20 mol$ of $Fe X^{3 +}$ until all iron is deposited as the metal. Use $E^{\circ} (Zn X^{2 +} / Zn) = - 0.76 V$ , $E^{\circ} (Fe X^{2 +} / Fe) = - 0.44 V$ , $E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = + 0.77 V$ and $Zn = 65$ . The mass of zinc consumed is

Zinc (

- 0.76 V

) can reduce

Fe X^{3 +}

Fe X^{2 +}

and also

Fe X^{2 +}

Fe

(since

- 0.44 > - 0.76

). Overall

2 Fe X^{3 +} + 3 Zn \to 2 Fe + 3 Zn X^{2 +}

, so

0.20 mol

Fe needs

0.30 mol

Zn, i.e.

0.30 \times 65 = 19.5 g

. Stopping at

Fe X^{2 +}

would need only

6.5 g