A current of 4.0 A is passed for 5 hours through 1L of 2M solution of nitrate using two nickel electrodes. The molarity of the solution at the end of the electrolysis will be:
1.2 M
2.0 M
1.5 M
2.5 M
A
1.5 M
B
2.0 M
C
1.2 M
D
2.5 M
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Solution
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Ni2++2e−⟶Ni
Faraday =4×5×60×6096500=0.746
∴ Moles of Ni reduced =0.7462=0.373mol
Moles of Ni2+ before reaction =2×1=2mol
Left out =2−0.373=1.626
∴Molarity=MolesVolume=1.6261=1.626molLit.
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