`30 mL` of `0.13 M NiSO_(4)` is electrolysed using a current of `360` milliamperes for `35.3` minutes. The mass of the metal that would have been plated out if current efficiency is only `60%` (`Ni = 58.7 u`) isA. `0.9131 g`B. `0.3911 g`C. `0.1391 g`D. `0.2474 g`

`30 mL` of `0.13 M NiSO_(4)` is electrolysed using a current of `360`

1.	`30 mL` of `0.13 M NiSO_(4)` is electrolysed using a current of `360` milliamperes for `35.3` minutes. The mass of the metal that would have been plated out if current efficiency is only `60%` (`Ni = 58.7 u`) isA. `0.9131 g`B. `0.3911 g`C. `0.1391 g`D. `0.2474 g`
Answer» Correct Answer - C The current efficiency is define as the ration of the amount of the converted substance to the amount of the substance that would react if the whole current is used in a given reaction. Thus `%` current efficiently `= ("Actuall mass of Ni deposited")/("Expected mass of Ni deposited") xx 100%` Calculate the mass of `Ni` that should be deposited : Charge in coulombs = current in amperes `xx` time in seconds `= (0.36 A)(35.3 "minutes")(60 "seconds/min")` `= 762.48 C` No. of faraday `= (762.48 C)/(96500 C//F)` `= 0.0079 F` Since `1 F` (1 mole of electric charge) produces `1` equivalent of the substance, no of equivalents of `Ni` that should be deposited for `100%` current efficiency is equal to no. of faradays. Thus, mass of `Ni` deposited for `100%` current efficiency will be (no of equivalents)(`gm` equivalent mass) `= (0.0079 eq.)(58.7//2 g eq.^(-1))[Ni^(2+)+2e^(-)rarrNi]` But the current effciency is `60%` only. Thus Actual mass of `Ni` deposited `= 0.2318 xx (60)/(100)` `= 0.1391 g`

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`30 mL` of `0.13 M NiSO_(4)` is electrolysed using a current of `360` milliamperes for `35.3` minutes. The mass of the metal that would have been plated out if current efficiency is only `60%` (`Ni = 58.7 u`) isA. `0.9131 g`B. `0.3911 g`C. `0.1391 g`D. `0.2474 g`

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