Question
What amount of current is required to be passed during a period of 5 minutes in order to deposit 0.6354gm of copper by electrolysis of aqueous cupric sulphatesolution? (Molar wt. of Copper = 63.5gm/mole)
Answer: Option B
:
B
Moles of copper to be obtained =0.635463.54=0.01 moles
Cu+++2e−→Cu
In order to obtain 1 mole of copper , 2 Felectricity is required .
so, to obtain 0.01 moles of copper, 0.02 F of electricity will be needed.
Q=I×t=0.02F=0.02×96500C=1930C of electricity
Or, I×(5×60)=1930
⇒I=19330=6.43A
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:
B
Moles of copper to be obtained =0.635463.54=0.01 moles
Cu+++2e−→Cu
In order to obtain 1 mole of copper , 2 Felectricity is required .
so, to obtain 0.01 moles of copper, 0.02 F of electricity will be needed.
Q=I×t=0.02F=0.02×96500C=1930C of electricity
Or, I×(5×60)=1930
⇒I=19330=6.43A
Was this answer helpful ?
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