The electrolysis of MnO₂ is carried out from a solution of MnSO₄ in H₂SO₄ (aq.). If a current of 25.5 ampere is used with a current efficiency of 85%, how long would it take to produce 1 kg of MnO₂?

Please explain briefly .

The overall reaction can be written as : Mn²⁺(aq) + 2H₂O(l)--> MnO₂(s) + 2H⁺(aq) + H₂(g)

So there is transfer of 2 e-.

formula used to calculate time w = z I t

w = wt of MnO2 deposited = 1kg = 1000 g.

z = electrochemical equivalent = equivalent wt. of MnO2 / 96500 = Mol. wt. of MnO2/2 /965000(eq. wt.of MnO2 = mol. wt. of MnO2 /2,since there is transfer of 2 electrons) = 86.9 /2/96500 = 0 .00045

I= 25.5 A but 85 % efficient ∴ 25.5 *85/100 = 21.675 A

Putting the value in formula to get t = w /z I = 1000g / 0 .00045/ 21.675 A = 1.025 x 10⁵ seconds.