Question

If equal quantities of electricity are passed through three voltmeter containing $FeS{O}_4$, $F{e}_2(S{O}_4{)}_3$ and $Fe(N{O}_3{)}_2$, then which of the following is not true?

• A. Amount of iron deposited in $FeS{O}_4$ and $F{e}_2(S{O}_4{)}_3$ is equal
• B. Amount of iron deposited in $Fe(N{O}_3{)}_3$ is $\frac{2}{3}$ of the amount of iron deposited in $FeS{O}_4$
• C. Amount of iron deposited in $F{e}_2(S{O}_4{)}_3$ and $Fe(N{O}_3{)}_3$ are equal
• D. Same gas will evolve in all three cases at anode

Answer 1

The correct option is A Amount of iron deposited in $FeS{O}_4$ and $F{e}_2(S{O}_4{)}_3$ is equal

(1) $FeS{O}_4\to F{e}^{2+}+S{O}_4^{2-}$

$F{e}^{2+}+2e^{-}\to F{e}_{\left(s\right)}$

$\therefore 2F$ charge is needed for $1$ mole deposition of $Fe$ .

(2) $F{e}_2(S{O}_4{)}_3\to 2F{e}^3+3S{O}_4^{2-}$

$F{e}^{3+}+3e^{-}\to F{e}_{\left(s\right)}$

$3F\to 1$ mole

$2F$ charge is needed for $\frac{2}{3}$ mole deposition of $Fe$.

The gas liberated in all these cases is the same and that is $H_2$ gas.

Therefore, option A is correct.