In $\text{CuS}{{\text{O}}_{\text{4}}}$solution when electric current equal to $2.5$faraday is passed, the gm equivalent deposited on the cathode is
A.$1$
B.$1.5$
C.$2$
D.$2.5$

Hint: According to faraday's law of electrolysis, one gram equivalent of a substance deposited is equal to one Faraday of charge passing during electrolysis. Thus to approach this question we have to use Faraday’s law of electrolysis to calculate the deposited copper on the cathode.

Formula used:
Gram-equivalent $=\dfrac{M}{{{n}_{f}}}$
M = molecular mass
${{n}_{f}}=$n-factor

Complete Step by Step Answer:
Here the electrolysis reaction occurs in $CuS{{O}_{4}}$solution. $C{{u}^{2+}}$ion from solution deposited as metallic$Cu$ on the cathode.
$C{{u}^{2+}}(aq.)+2{{e}^{-}}\to Cu(s)$
The charge required for the decomposition of one mole of copper $=2F$
$1$mole $Cu$$=63.5$g $Cu$
n-factor = $2$ = one mole $Cu$is deposited by using $2$mole electrons
Hence $1$gm-equivalent of $Cu$$=\dfrac{M}{{{n}_{f}}}=\dfrac{63.5}{2}g$
From Faraday’s law of electrolysis,
$2F=1$mole $Cu(s)=63.5g$ of $Cu(s)$
$\therefore \dfrac{63.5}{2}g$of $Cu(s)$ $=\dfrac{2}{63.5}\times \dfrac{63.5}{2}=1F$
Or, $1$gm-equivalent of $Cu$$=1F$
Hence $2.5$ $F$$=2.5$ gm-equivalent of $Cu$
Therefore $2.5$gm-equivalent $Cu$is deposited on the cathode.
Thus, option (D) is correct.

Additional information: Faraday’s law of electrolysis explains the relationship between the amount of electric charge passed through an electrolyte and the amount of deposited substance at any electrode. Faraday’s first law tells us chemical deposition caused by the flow of a current is directly proportional to the amount of electricity passing through an electrolyte. Whereas Faraday’s second law states that the amount of deposited or liberated chemical at any electrode on passing a certain amount of current is directly proportional to their equivalent weight.

Note: The charge of one Faraday is equal to $96500$ coulomb, therefore one Faraday is defined as the amount of charge on one mole of electrons. We should remember the value of one Faraday $96500$ coulomb/mole as this value is used in many problems of electrolysis reaction.