JEE MAIN - Chemistry (2019 - 9th April Evening Slot - No. 8)
A solution of Ni(NO3)2 is electrolysed between
platinum electrodes using 0.1 Faraday
electricity. How many mole of Ni will be
deposited at the cathode?
0.10
0.15
0.20
0.05
Explanation
Cathode reaction :
Ni+2 + 2e- $$ \to $$ Ni(s)
$$ \therefore $$ From 2 mole of electrons 1 mole of Ni is deposited at the cathode.
So from 0.1 F or 0.1 mole of electrons $${1 \over 2} \times 0.1$$ = 0.05 mole of Ni is deposited at the cathode.
Ni+2 + 2e- $$ \to $$ Ni(s)
$$ \therefore $$ From 2 mole of electrons 1 mole of Ni is deposited at the cathode.
So from 0.1 F or 0.1 mole of electrons $${1 \over 2} \times 0.1$$ = 0.05 mole of Ni is deposited at the cathode.
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