You visited us 2 times! Enjoying our articles? Unlock Full Access!

Electrolysis and Electrolytes

A current of ...

Question

A current of $3.7$ ampere is passed for 6 hrs. between Ni electrodes in $0.5$ litre of $2 M$ solution of ${N i ({N O}_{3})}_{2}$ What will be the molarity of solution at the end of electrolysis?

Open in App

Solution

$I n i t i a l M = 2 M o l a r$ $M o (N i) = 58.7 g M o l^{- 1}$
$M o l a r i t y f i n a l = 2 M - ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \frac{58.7}{2}      E \times \frac{3.7 \times 6 \times 3600}{96500}      F \times (\frac{1}{58.7 \times 0.5}) ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭$
$= 1.171 M o l a r$

Suggest Corrections

Similar questions

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

N i

2 M

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

3.7

A

0.5 L

2 M

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

2 \times 10^{- 3} M

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

Join BYJU'S Learning Program