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How many moles of Ni will be deposited at the cathode if solution of \mathrm{Ni}\left(\mathrm{NO}_3\right)_2 is electrolysed between platinum electrode using 4.25 faraday.

Option: 1

1.2 mol


Option: 2

2.1 mol


Option: 3

3.3 mol


Option: 4

4.2 mol


Answers (1)

best_answer

\mathrm{Ni}\left(\mathrm{NO}_3\right)_2 \rightarrow \mathrm{Ni}^{2+}+2 \mathrm{NO^{-}}_3
2 F of Current deposits \rightarrow 1 mole
\text { 4. } 25 \text { of current deposits }=\frac{4.25}{2}=2.1

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