Equal quantities of electricity are passed through 3 voltameters containing ${FeSO}_{4}, {Fe}_{2} {({SO}_{4})}_{3} and Fe {({NO}_{3})}_{3} .$
Consider the following statement .
(A) The amounts of iron deposited in ${FeSO}_{4} and {Fe}_{2} ({SO}_{4})_{3}$
are equal .
(B) The amount of iron deposited in Fe(NO₃)₃ is ⅔ rd of the amount deposited in FeSO₄
(C) The amount of iron deposited in ${Fe}_{2} {({SO}_{4})}_{3}$ and $Fe {({NO}_{3})}_{3}$ are equal.

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(B) is correct

(A) is correct

Both (A) and (B) are correct

answer is B, C.

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Detailed Solution

Suppose $2 F$ of electricity passed through ${FeSO}_{4},$ ${Fe}_{2} {({SO}_{4})}_{3} and Fe {({NO}_{3})}_{3}$ solution

$\begin{array}{l} {FeSO}_{4} \to {Fe}^{2 +} + {SO}_{4}^{2 -} \\ {Fe}^{2 +} + 2 e^{-} \to Fe (S) \\ 2 F 1 mole \\ {Fe}_{2} {({SO}_{4})}_{3} \to 2 {Fe}^{3 +} 3 {SO}_{4}^{2 -} \\ {Fe}^{3 +} + 3 e^{-} \to Fe (S) \\ 3 F 1 mole \\ ∴ 2 F will give = \frac{2}{3} mole \\ Fe {({NO}_{3})}_{3} \to {Fe}^{3 +} + 3 {NO}_{3}^{-} \\ {Fe}^{3 +} + 3 e^{-} \to Fe (S) \\ 3 F 1 mole \\ ∴ 2 F will give = \frac{2}{3} mole \end{array}$