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Question
A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
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Solution
The reaction takes place in the following manner:
\[\ce{Ni^{2+} + 2e- -> Ni}\]
Atomic weight of \[\ce{Ni}\] = 58.70
Equivalent weight of Ni = \[\ce{\frac{Atomic weight}{Number of valence electrons}}\]
= \[\ce{\frac{58.70}{2}}\]
= 29.35
According to Faraday’s first law of electrolysis,
\[\ce{W = Z . I . t = \frac{Equivalent weight}{96500} \times I \times t}\]
= \[\ce{\frac{29.35}{96500} × 5 \times 20 \times 60}\]
= 1.825 g
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