Question

Equal quantities of electricity are passed through three voltameters containing $FeS{O}_4,F{e}_2(S{O}_4{)}_3$, and $Fe(N{O}_3{)}_3$. consider the following statements in this regard
(1) The amount of iron deposited in $FeS{O}_4$ and $Fe(S{O}_4{)}_3$ is equal
(2) The amount of iron deposited in $Fe(N{O}_3{)}_3$ is two thirds of the amount of iron deposited in $FeS{O}_4$
(3) The amount of iron deposited in $F{e}_2\left(S{O}_4{)}_{3}Fe\right(N{O}_3{)}_3$ is equal
Of these statements

• A. 1 and 2 are correct
• B. 2 and 3 are correct
• C. all are correct
• D. none of the above

Answer 1

The correct option is A 1 and 2 are correct

According to Faraday's law of electrolysis , weight of any element deposited is directly proportional to its equivalent weight.

In Fe${SO}_4$ , iron is in +2 oxidation state , when it will get deposited then it will come to 0 oxidation state(Fe form) , hence change in oxidation number = 2.

So its equivalent wt = wt of Fe /2

In$Fe({NO}_3{)}_3$ , iron will change its oxidation number from +3 to 0 , hence equivalent wt of iron in this case = wt of Fe/3

Similarly in ${Fe}_2$$({SO}_4{)}_3$ , Iron is in +3 oxidation state , and it will change its oxidation state from +3 to 0 , so equivalent weight of iron in this case = wt of Fe / 3

Since equivalent weight of $Fe({NO}_3{)}_3$ & ${Fe}_2$$({SO}_4{)}_3$ is equal , So their amount deposited is equal .

Also comparing the equivalent weight we get that the amount of iron deposited in $Fe({NO}_3{)}_3$ is 2/3 times the amount deposited in Fe${SO}_4$

Correct Option is B

Fe(NO3s two thirds of the amount of iron deposited