Question

If same quantity of electricity is passed through three electrolytic cells containing $FeS{O}_4,F{e}_2(S{O}_4{)}_3$, and $Fe(N{O}_3{)}_3$, then:

• A. the amount of iron deposited in $FeS{O}_4$ and $F{e}_2(S{O}_4{)}_3$ are equal.
• B. the amount of iron deposited in $FeS{O}_4$ is 1.5 times of the amount of iron deposited in $Fe(N{O}_3{)}_3$.
• C. the amount of iron deposited in $Fe(S{O}_4{)}_3$ and $Fe(N{O}_3{)}_3$ are equal.
• D. the same amount of gas is evolved in all three cases of the anode.

Answer 1

Answer: (B), (C), (D)

The correct options are
B the amount of iron deposited in $FeS{O}_4$ is 1.5 times of the amount of iron deposited in $Fe(N{O}_3{)}_3$.
C the same amount of gas is evolved in all three cases of the anode.
D the amount of iron deposited in $Fe(S{O}_4{)}_3$ and $Fe(N{O}_3{)}_3$ are equal.

The options (B), (C) and (D) are correct. The option (A) is incorrect.

(B) The amount of iron deposited in $FeS{O}_4$ is 1.5 times the amount of iron deposited in $Fe(N{O}_3{)}_3$.

(C) The amount of iron deposited in $Fe(S{O}_4{)}_3$ and $Fe(N{O}_3{)}_3$ are equal.

(D) The same amount of gas is evolved in all three cases of the anode.
$\begin{array}{ccc}& & \\ FeS{O}_4;& F{e}^{2+}+2e^{-}\to Fe& 1F\equiv \frac{1}{2}molFe\end{array}$

$\begin{array}{cccc}F{e}_2(S{O}_3{)}_3:& F{e}^{3+}+3e^{-}\to Fe& \Rightarrow & 1F\equiv \frac{1}{3}molFe\end{array}$

$\begin{array}{ccc}Fe(N{O}_3{)}_3;& F{e}^{3+}+3e^{-}\to Fe& 1F\equiv \frac{1}{3}molFe\end{array}$

Amount of Fe deposited in $F{e}_2(S{O}_4{)}_3=\frac{1}{2}:\frac{1}{3}=1.5:1$
At anode : $4\stackrel{\ominus }{O}H\to O_2+2H_{2}O+4e^{-}$

In all cases, the same amount of gas is evolved at the anode.