A current of $3.7$ $A$ is passed for 6 h between nickel electrodes in $0.5 L$ of a $2 M$ solution of $N i (N O_{3})_{2} .$ The molarity (M) of the solution at the end of electrolysis will be: ( write your answer to the nearest integer)

Open in App

Solution

Given, I = 3.7 A
$t = 6 \times 60 \times 60 s$
Q = I.t
$= 3.7 \times 6 \times 60 \times 60 C$ = 79920 C
$∵$ 96500 C of electricity deposit Ni = 1 Eq
$∴$ 79920 C of electricity will deposit $N i = \frac{1 \times 79290}{96500}$
= 0.828 Eq
Number of moles of $N i = \frac{E q o f N i}{V a l e n c y} = \frac{0.828}{2}$ = 0.414 mol
[ $∵$ In $N i (N O_{3})_{2}$ , valency of Ni = 2]
Volume (V) = 0.5 L
Molarity (M) = 2 M
Molarity = $\frac{N u m b e r o f m o l e s}{V o l u m e (i n L)}$
$2 = \frac{N u m b e r o f m o l e s}{0.5}$
Number of moles = 1.0 mol
Thus, nickel left in 0.5 L of solution = Number of moles of Ni in original solution - Number of moles of Ni deposited
= 1.0 - 0.414
Thus, molarity (M) = $\frac{0.586}{0.5}$ = 1.172 M

Suggest Corrections

Similar questions

N i (N O_{3})_{2}

3.7

0.5

2 M

{N i ({N O}_{3})}_{2}

3.7

6

0.5

2 M

N i (N O_{3})_{2}

(1 F = 96500 c o u l o m b; N i = 58.7

2 \times 10^{- 3} M

N i {({N O}_{3})}_{2}

N i

2 M

N i (N O_{3})_{2}

Join BYJU'S Learning Program