Question

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

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

Answer 1

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

$FeS{O}_4:F{e}^{2+}+2e^{-}\to Fe1F\equiv \frac{1}{2}molFe$ $Fe{\left(N{O}_3\right)}_3:F{e}^{3+}+3e^{-}\to Fe1F\equiv \frac{1}{3}molFe$ $F{e}_2{\left(S{O}_3\right)}_3:F{e}^{3+}+3e^{-}\to Fe\Rightarrow 1F\equiv \frac{1}{3}molFe$
Amount of $Fe$ deposited in $Fe{\left(N{O}_3\right)}_3=$ Amount of $Fe$ deposited in $F{e}_2{\left(S{O}_4\right)}_3$
Amount of $Fe$ deposited in $F{e}_2{\left(S{O}_4\right)}_3=\frac{1}{2}:\frac{1}{3}=1.5:1$
At anode $:4\stackrel{\ominus }{\mathrm{OH}}⟶O_2+2H_{2}O+4e^{-}$
In all cases same amount of gas is evolved at the anode