Question

Equal quantities of electricity are passed through 3 voltameters containing $FeSO_{4}\, Fe_{2}\,\left(SO_{4}\right)_{3}$ and $Fe\left(NO_{3}\right)_{3^{-}}$ Consider the following statement:
(1)The amounts of iron deposited in $FeSO_{4}$ and $Fe_{2} \left(SO_{4}\right)_{3}$ are equal
(2)The amount of iron deposited in $Fe\left(NO_{3}\right)_{3}$ is 2/3rd of the amount deposited in $FeSO_{4}$
(3)The amount of iron deposited in $Fe\left(SO_{4}\right)_{3}$ and $Fe\left(NO_{3}\right)_{3}$ are equal

• A. (1) is correct
• B. (2) is correct
• C. (3) is correct
• D. Both (1) and (2) are correct

Answer 1

Answer: (B), (D)

Suppose 2F of electricity passed through $FeSO_{4},\, Fe_{2} \left(SO_{4}\right)_{3}\, and \, Fe\left(NO_{3}\right)_{3}$ solution
$FeSO_{4}\to Fe^{2+}+SO_{4}^{2-}$
$Fe^{2+}+^{2e-}_{2F}\to^{Fe\left(S\right)}_{1 mole} $
$Fe_{2} \left(SO_{4}\right)_{3}\,\to2Fe^{3+}+3SO^{2-}_{4}$
$Fe^{3+}+^{3e^-}_{3F}\to^{Fe\left(S\right)}_{1 mole}$
$\therefore 2F will \, give =\frac{2}{3} mole$
$Fe\left(NO_{3}\right)_{3} \to Fe^{3+}+3NO^{-}_{3}$
$Fe^{3+} +^{3e^-}_{3F} \to^{Fe\left(S\right)}_{1 mole}$
$\therefore 2F\, will \, give =\frac{2}{3} mole$