A solution of $N i (N O_{3})_{2}$ is electrolyzed between platinum electrodes using $0.1 F a r a d a y$ electricity. Moles of $N i$ that will be deposited at the cathode:

0.20

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0.10

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0.15

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0.05

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Solution

The correct option is D $0.05$
A solution of $N i (N O_{3})_{2}$ is electrolysed between platinum electrodes using $0.1 F$ electricity. It means that $0.1$ equivalent of $N i^{2 +}$ will be discharged.
Electrolysis of $N i (N O_{3})_{2}$ gives:
$N i^{2 +}$ + $2 e -$ $\to$ $N i$
(Atomic mass of $N i = 58.7)$
Number of equivalents = Number of moles $\times$ number of electrons.
⇒ $0.1$ = Number of moles $\times 2$
= Number of moles of $N i = \frac{0.1}{2} = 0.05$

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A solution of Ni(NO3)2 is electrolyzed between platinum electrodes using 0.1 Faraday electricity. Moles of Ni that will be deposited at the cathode:

A solution of $N i (N O_{3})_{2}$ is electrolyzed between platinum electrodes using $0.1 F a r a d a y$ electricity. Moles of $N i$ that will be deposited at the cathode: