1.

If 2 faraday current is pass through CuSO_(4) solution then find out how much copper can be deposited on cathode ? [Cu=63.5 gm "mole"^(-1)]

Answer»

0 gm
2 gm
63.5 gm
127 gm

Solution :`CuSO_(4)toCu_((aq))^(2+)+SO_(4(aq))^(2-)`. . . Ionization
`Cu_((aq))^(2+)+2E^(-) to Cu_((S))` . . . (Cathodic reduction)
2 mole `e^(-) to ` 1 MOL COPPER
So 2F current produce 63.5 gm copper


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