Question

If same quantity of electricity is passed through three electrolytic cells containing $FeSO _4, Fe _2\left( SO _4\right)_3$, and $Fe \left( NO _3\right)_3$, then

• A. The amount of iron deposited in $FeSO _4$ and $Fe _2\left( SO _4\right)_3$ are equal
• B. The amount of iron deposited in $FeSO _4$ is $1.5$ times of the amount of iron deposited in $Fe \left( NO _3\right)_3$
• C. The amount of iron deposited in $Fe _2\left( SO _4\right)_3$ and $Fe \left( NO _3\right)_3$ are equal
• D. The same amount of gas is evolved in all three cases of the anode

Answer 1

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

$FeSO _4: Fe ^{2+}+2 e^{-} \rightarrow Fe 1 F \equiv \frac{1}{2} mol Fe$ $Fe \left( NO _3\right)_3: Fe ^{3+}+3 e^{-} \rightarrow Fe 1 F \equiv \frac{1}{3} mol Fe$ $Fe _2\left( SO _3\right)_3: Fe ^{3+}+3 e^{-} \rightarrow Fe \Rightarrow 1 F \equiv \frac{1}{3} mol \,Fe$
Amount of $Fe$ deposited in $Fe \left( NO _3\right)_3=$ Amount of $Fe$ deposited in $Fe _2\left( SO _4\right)_3$
Amount of $Fe$ deposited in $Fe _2\left( SO _4\right)_3=\frac{1}{2}: \frac{1}{3}= 1.5: 1$
At anode $: 4 \overset{\ominus}{ OH } \longrightarrow O _2+2 H _2 O +4 e^{-}$
In all cases same amount of gas is evolved at the anode