Question

A current of 250. A flows for 24.0 hours at an anode where the reaction occurring is       Mn2+(aq) + 2H2O(l) → MnO2(s) + 4H+(aq) + 2e– What mass of MnO2 is deposited at this anode?

		19.5 kg
		4.87 kg
		2.43 kg
		12.9 kg
		none of the above

Answer 1

Answer:

If the current is 250 A, it is also 250 C/s. This way we can find how many coulombs of electrons have passed through the anode if it was allowed to flow for 24.0 hrs.

24.0 hr * 60 min/ 1 hr * 60 s/ 1 min * 250 C/s = 2.16*10^7 C

Using Faraday's constant, we know that for every 96,485 C, there is 1 mol e^-. So now we can see how many mols of electrons have been passed through.

2.16*10^7 C * 1 mol e^-/96485 C = 223.87 mol e^-

Using the balanced equation, since we know how many mols of electrons are present in the anode solution, we can see how many mols of MnO₂ are deposited on the anode (by using a molar ratio):

Mn²⁺(aq) + 2H₂O(l) → MnO₂(s) + 4H⁺(aq) + 2e^-

223.87 mol e^- * 1 mol MnO₂/ 2 mol e^- = 111.93 mol MnO₂

Using the molar mass of manganese dioxide (54.94 g/mol + (2* 15.9994 g/mol) = 86.94 g/mol), we can find the mass of MnO₂.

111.93 mol MnO₂ * 86.94 g MnO2/ 1 mol MnO2 = 9.73*10^3 g MnO2, or 9.73 kg MnO₂has been desposited at the anode.