Question

Equal quantities of electricity are passed through 3 voltameters containing $FeS{O}_{4}F{e}_2{\left(S{O}_4\right)}_3$ and $Fe{\left(N{O}_3\right)}_{3^{-}}$ Consider the following statement:
(1)The amounts of iron deposited in $FeS{O}_4$ and $F{e}_2{\left(S{O}_4\right)}_3$ are equal
(2)The amount of iron deposited in $Fe{\left(N{O}_3\right)}_3$ is 2/3rd of the amount deposited in $FeS{O}_4$
(3)The amount of iron deposited in $Fe{\left(S{O}_4\right)}_3$ and $Fe{\left(N{O}_3\right)}_3$ 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 $FeS{O}_4,F{e}_2{\left(S{O}_4\right)}_{3}andFe{\left(N{O}_3\right)}_3$ solution
$FeS{O}_4\to F{e}^{2+}+S{O}_4^{2-}$
$F{e}^{2+}{+}_{2F}^{2e-}{\to }_{1mole}^{Fe\left(S\right)}$
$F{e}_2{\left(S{O}_4\right)}_3\to 2F{e}^{3+}+3S{O}_4^{2-}$
$F{e}^{3+}{+}_{3F}^{3e^{-}}{\to }_{1mole}^{Fe\left(S\right)}$
$\therefore 2Fwillgive=\frac{2}{3}mole$
$Fe{\left(N{O}_3\right)}_3\to F{e}^{3+}+3N{O}_3^{-}$
$F{e}^{3+}{+}_{3F}^{3e^{-}}{\to }_{1mole}^{Fe\left(S\right)}$
$\therefore 2Fwillgive=\frac{2}{3}mole$