Question

One litre of 1 M $CuS{{O}_{4}}$ solution is electrolysed. After passing 2F charge, the molarity of $CuS{{O}_{4}}$ will be ________________.
A. M/2
B. M/4
C. M
D. Zero

Answer 1

Hint: Molarity is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by the letter M. n – factor in redox reaction the substance factor is equal to the number of moles of electron lost or gained per molecule. A balanced equation is an equation for a chemical reaction in which both the reactants and the products have the same number of atoms for each element in the reaction, and the total charge.

Complete answer:
The given data is as follow:
Charge, (q) = 2F, volume of solution = 1L, molarity (before electrolysis) = 1M
 It is known that the relation between charge, valence factor and moles is as follows.
$Moles\times Valence Factor=\dfrac{q}{F}$ …(I)
Moles = Molarity (initial or before electrolysis) $\times $ Volume
Molarity is defined as the number of moles of solute dissolved in one litre (or one cubic decimeter) of solution. It is denoted by M.
Molarity (before electrolysis) given in question = 1M
Volume given in question = 1L.
Moles = $1M\times 1L$ = 1 moles.
Valence factor is also known as n-factor. The N-factor of substance in redox reaction is equal to the number of moles of lost or gained electrons per molecule. N-factor of substance in non – redox reaction is equal to the product of displaced mole and its charge.
The reaction equation will be followed as:
$C{{u}^{2+}}+2e\to Cu$
Hence, the n-factor or valence factor of the reaction is 2.
Now, put all value in equation (I),
$1\times 2=\dfrac{q}{F}$
q = 2F
therefore, passing 2F charge would consume the solution completely.
Hence, the molarity of the solution after passing 2F charge is 0.

Hence, option D is correct.

Note:
Don’t get confused between molarity and molality. Molarity (M) is defined as the number of moles of solute dissolved in one litre of solution whereas Molality (m) is defined as number of moles of solute per kilogram (kg) of the solvent.
To avoid calculation mistakes, convert all quantities into SI units before calculation.