First slide

Faraday's Laws

Question

When a quantity of electricity is passed through CuSO₄ solution 0.16g of copper gets deposited. If the same quantity of electricity is passed through acidulated water. Then the volume of H₂ liberated at STP will be [given : at.wt.of Cu = 64]

Moderate

Solution

According to 2nd law of Faraday for same Faradays of charge consumed same number of equivalents are produced.
$Number\,\,of\,\,\,equivalents\,\,\,deposited = \frac{{given\,mass}}{{Equivalent\,\,mass}}$
$Number\,\,of\,\,\,equivalents\,\,\,of\,Cu\,\,deposited = Number\,of\,equivalents\,\,\,of\,\,\,{H_2}\,liberated = \frac{{0.16}}{{32}} = 0.005$
$Volume\,of\,{H_2}\,liberated\,at\,STP = Number\,of\,equivalents\,liberated \times Volume\,equivalent$
$Volume\,of\,{H_2}\,liberated\,at\,STP = 0.005 \times 11200 = 56c{m^3}$

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