Q8 of 26 Page 1

Calculate the time to deposit 1.17 g of Ni at cathode when a current of 5 A was passed through the solution of Ni(NO3)2. (Molar mass of Ni = 58.5 g/mol, 1 F = 96500 C/mol)

Given, mass of Ni deposited (w) = 1.17 g


Molar mass of Ni (M) = 58.5 g/mol


Current (I) = 5A


Number of mole of Ni deposited (n) = = 0.02 moles


Ni2+ (aq) + 2e⁻ → Ni(s)


We know, charge of 1mole of electron is 1F i.e. 96500 C/mol. Thus, Ni2+ requires (2F 0.02) C of charge to deposit 0.02 mole of Ni.


So, Q = It


Therefore, t = =


= 772 sec = 12.86 min


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