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A current of 3.7 ampere is passed for 6 hr between Ni electrod

Question

A current of 3.7 ampere is passed for 6 hr 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?

A

1.172 M

B

2M

C

3.2M

D

None of these

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A current of 3.7 ampere is passed for 6 hr 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?

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Text solutionVerified

The electrode is given as:

N i^{2 +} + 2 e^{-} \to N i

For 1 mole of electrolysis, we require 2x96500C

The quantity of the charge passed can be given as

= 3.7 \times 6 \times 60 \times 60 co l o u mb = 79920 co l o u mb

Number of moles of nickel in the electrolysis

= \frac{79920}{2 \times 96500} = 0.414 m o l

Thus, number of moles of nickel deposited

= (2 \times 0.5) - 0.414 = 0.586 m o l

Thus, 0.586 mol of Ni are present in 0.5 litre,

Thus, molarity of the left 2 M solution

= \frac{n}{V} = \frac{0.586}{0.5} M = 1.172 M

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Question Text	A current of 3.7 ampere is passed for 6 hr 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?
Answer Type	Text solution:1
Upvotes	150