Electrochemistry 1 Question 28
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2. A solution of $\mathrm{Ni}\left(\mathrm{NO}{3}\right){2}$ 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)
======= ####2. A solution of $\mathrm{Ni}\left(\mathrm{NO}{3}\right){2}$ 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)
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) 0.20
(b) 0.10
(c) 0.15
(d) 0.05
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Answer:
Correct Answer: 2. (b)
Solution:
- A solution of $\mathrm{Ni}\left(\mathrm{NO}{3}\right){2}$ 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 $\mathrm{Ni}\left(\mathrm{NO}{3}\right){2}$ 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 $\mathrm{Ni}=\frac{0.1}{2}=0.05$
