Electrochemistry 1 Question 2
2. A solution of $\mathrm{Ni}(\mathrm{NO} _{3}) _{2}$ is electrolysed between platinum electrodes using $0.1$ Faraday electricity. How many mole of $\mathrm{Ni}$ will be deposited at the cathode?
(2019 Main, 9 April II)
(a) $0.20$
(b) $0.10$
(c) $0.15$
(d) $0.05$
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Answer:
Correct Answer: 2. (d)
Solution:
- A solution of $\mathrm{Ni}(\mathrm{NO} _{3}) _{2}$ is electrolysed between platinum electrodes using $0.1$ Faraday electricity. It means that $0.1$ equivalent of $\mathrm{Ni}^{2+}$ will be discharged.
Electrolysis of $\mathrm{Ni}(\mathrm{NO} _{3}) _{2}$ gives
$ \mathrm{Ni}^{2+}+2 e^{-} \longrightarrow \mathrm{Ni} \quad(\text { Atomic mass of } \mathrm{Ni}=58.7) $
Number of equivalents $=$ Number of moles $\times$ number of electrons.
$0.1=$ Number of moles $\times 2$
$\therefore$ Number of moles of $\mathrm{Ni}=\frac{0.1}{2}=0.05$
BETA
