Question

The molar mass of $CuS{O_4}.5{H_2}O$is 249. Its equivalent mass in the reaction (a) and (b) would be:
(a) Reaction of $CuS{O_4} + KI \to Product$
(b) Electrolysis of $CuS{O_4}.5{H_2}O$ solution.

A) (a) 249 (b) 249
B) (a) 124.5 (b) 124.5
C) (a) 249 (b) 124.5
D) (a) 124.5 (b) 249

Answer 1

Hint: Recall the Faraday's second law of electrolysis which states that the amount of substance liberated when the electricity is passed through the electrolytic solution are proportional to their chemical equivalent weights.
Chemical equivalent weight = atomic mass of metal / no. of electrons required to reduce the cation.


Complete step by step answer:
Here, in part (a) : reaction of copper sulphate with KI taking place.
$\begin{gathered}
CuS{O_4} + KI \to Product
CuS{O_4}.5{H_2}O + KI \to C{u_2}{I_2} + {I_2} + {K_2}S{O_4} + {H_2}O

\end{gathered} $
In this, the oxidation state of copper in $CuS{O_4}$ is +2 and the oxidation state of copper in $C{u_2}{I_2}$ is +1. Oxidation state of copper is changing from +2 to +1 ,hence 2 electrons needed to reduce the copper .

Since, ${\text{equivalent weight = }}\dfrac{{{\text{atomic mass}}}}{{{\text{no}}{\text{. of electrons required to reduce the cation}}{\text{.}}}}$ $\begin{gathered}
 {\text{equivalent weight }} = \dfrac{{249}}{2} = 124.5

\end{gathered} $
Therefore, $\begin{gathered}
 {\text{equivalent weight }} = \dfrac{{249}}{2} = 124.5
\end{gathered} $

(b) part: electrolysis of copper sulphate solution –
During electrolysis of copper, reduction of copper takes place on cathode .
Cathode: $C{u^{ + 2}} + 2{e^ - } \to Cu$
It is clear that 2 electrons are involved in the reaction to reduce copper.
Therefore,
 ${\text{equivalent weight = }}\dfrac{{{\text{atomic mass}}}}{{{\text{no}}{\text{. of electrons required to reduce the cation}}{\text{.}}}}$
     So again answer comes out to be 124.5
So, the correct answer is “Option B”.

Note: Michael faraday was the scientist who described the quantitative aspects of electrolysis. His 2 laws are:
1) First law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

2) Second law: the amount of substances liberated when the same quantity of electricity is passed through the electrolytic solution are proportional to their chemical equivalent weights.
And, ${\text{equivalent weight = }}\dfrac{{{\text{atomic mass}}}}{{{\text{no}}{\text{. of electrons required to reduce the cation}}{\text{.}}}}$