An aqueous solution of platinum salt was electrolysed by passing an electric current of 2.5 A for 2 hours. The number of moles of platinum metal deposited at the cathode was 0.0459 mol. Therefore, the charge of the platinum ion in the aqueous solution is ____?
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Given - number of moles - 0.0459 mol
Current - 2.5 A
Time - 2 hours
Find - charge
Solution - To find charge of platinum ion deposited, the formula used is -
Moles = I t / n F, where n is moles of electrons, f is Faraday's constant.
Charge represented by q will be calculated by the formula -
Q = it, where I is current and t is time
Q = 2.5*2
Q = 5 C
Moles = I t / n F
0.0459 = 5/n*96500
n = 108.9/96500
n = 0.0011
Charge of 0.0011 mole of electrons is - 1.602*10^-19*6.023*10^23*0.0011
= 106 C
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