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A current of 3.7 amper is passed for 6 hre between Pt electrodes in 0.5 litre, 2M solution of Ni(NO_(3))_(2). What will be the molarity of solution at the end of electrolysis?

Answer»

Solution :Charge in coulombs = current in amp `xx` time in s
`= 3.7 xx 6 xx 60 xx 60`
= 79920.
`THEREFORE` MOLE (faraday) of electric charge `= (79920)/(96500)`
`= 0.8283 F`.
Eq. of `Ni(NO_(3))_(2)` deompoased on electrolysis = 0.8283.
`therefore` m.e. of `Ni(NO_(3))_(2)` decomposed `= 0.8283 xx 1000 = 828.3`
Now, m.e. of `Ni(NO_(3))_(2)` solution before electrolysis
= NORMALITY `xx` volume (mL)
`= 4 xx 500 = 200`
{normality `= 2 xx` molarity `= 2 xx 2 = 4`}
m.e. of `Ni(NO_(3))_(2)` after electrolysis = 2000 - 828.3 = 1171.7
`therefore` molarity `Ni(NO_(3))_(2)` solution after electrolysis `= (2.34)/(2) = 1.17 M`.


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