Question

Two electrolytic cells one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. The ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells will be _____.
A:$3:1$
B: $2:1$
C:$1:1$
D:$3:2$

Answer 1

Hint: Equivalent weight is the ratio of atomic weight of the compound to the valency factor of that compound as in this case valency factor of iron in ferric chloride is +3 and valency factor of iron in ferrous chloride is +2 .

Complete answer:
$\
  {\text{Given,}} \\
  {\text{FeC}}{{\text{l}}_3}{\text{ and FeC}}{{\text{l}}_2} \\
  {\text{FeC}}{{\text{l}}_3} \to {\text{F}}{{\text{e}}^{ + 3}}{\text{ + 3C}}{{\text{l}}^{ - 1}} \\
  {\text{FeC}}{{\text{l}}_2} \to {\text{F}}{{\text{e}}^{ + 2}} + 2{\text{C}}{{\text{l}}^{ - 1}} \\
  {\text{So, FeC}}{{\text{l}}_3}{\text{ has F}}{{\text{e}}^{ + 3}}{\text{ ions}} \\
  {\text{ FeC}}{{\text{l}}_2}{\text{ has F}}{{\text{e}}^{ + 2}}{\text{ ions}} \\
  {\text{At, cathode iron deposition will take place;}} \\
  {\text{Now, calculating the equivalent weight of Iron,}} \\
  {\text{Equivalent weight of F}}{{\text{e}}^{ + 3}} = \dfrac{{{\text{Atomic weight of Iron}}}}{{{\text{Valency factor of F}}{{\text{e}}^{ + 3}}}} = \dfrac{{\text{M}}}{3} \\
  {\text{Equivalent weight of F}}{{\text{e}}^{ + 2}} = \dfrac{{{\text{Atomic weight of Iron}}}}{{{\text{Valency factor of F}}{{\text{e}}^{ + 2}}}} = \dfrac{{\text{M}}}{2} \\
  {\text{Ratio = }}\dfrac{{{\text{Equivalent weight of F}}{{\text{e}}^{ + 2}}}}{{{\text{Equivalent weight of F}}{{\text{e}}^{ + 3}}}} = \dfrac{{\dfrac{{\text{M}}}{2}}}{{\dfrac{{\text{M}}}{3}}} = \dfrac{3}{2} \\
\ $


Hence Option D is the correct answer.

Additional Information:
Calculation of Valency factor is basically done in 4 ways:
Type 1:
For acids if the only acid is given then you have to check the replaceable hydrogen atom it contains.if a reaction is given then you have to check the replaced hydrogen atom.
Type 2:
For bases if the only the acid is given then you have to check the replaceable hydroxyl ion it contain.if a reaction is given then you have to check the replaced hydroxyl ion
Type 3:
For salts you have to check the total positive charge on the salt .Here you don’t have to consider water .
Type 4:
For redox equation you have to consider the change in oxidation number and multiply with the number of atoms of the elements present in compound

Note:
Students should make sure that they don’t commit mistakes while calculating n-factor or valency factor.
Positive charge deposition takes place at cathode while negative charge deposition takes place at anode.