Electrolysis of water takes place as follows H2O⇌
H+
cathode
+
OH−
anode
At anode OH−
Oxidation
→
OH+e− 4OH→2H2O+O2 At cathode 2H++2e−
Reduction
→
H2 Given, time, t=1930s Number of moles of hydrogen collected =
1120×10−3
22.4
moles =0.05 moles ∵1 mole of hydrogen is deposited by =2 moles of electrons ∴0.05 moles of hydrogen will be deposited by =2×0.05 =0.10 mole of electrons Charge, Q=nF =0.1×96500 Charge, Q=it 0.1×96500=i×1930 i=