Current of 250. a flows for 24.0 hours at an anode where the reaction occurring is as follows: mn2 + (aq) + 2h2o (l) → mno2 (s) + 4h + (aq) + 2e-what mass of mno2 is deposited at this anode?

From Faraday's 1st law of electrolysis,

Total electricity passed into system = Q = IT = 250 X 24 X 60 X 60

= 2.16 X 10^7 C

We know that, 96500 C = 1 F

∴ 2.16 X 10^7 C = 223.8 F

Now, number of moles of MnO2 deposited = 223.8/2=111.9

Finally, 1 mole of MnO2 ≡ 86.94 g

∴ 111.9 mole of MnO2 ≡ 111.9 X 86.94 = 9728 g

Thus, mass of MnO2 that will be deposited at anode = 9728 g