A 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?

Question

Quizplus · Accepted Answer

To calculate the mass of MnO₂ deposited, we need to first calculate the amount of charge that has passed through the cell using Faraday's law: Q = I*t Q = (250 A) * (24.0 h) * (3600 s/h) = 21,600,000 C Next, we can use the stoichiometry of the reaction to calculate the moles of MnO₂ deposited for every 2 electrons: 2 mol e- -> 1 mol MnO₂ The total number of electrons passed through the cell is: n = Q/F n = (21,600,000 C) / (96,485 C/mol) = 224.01 mol e- Therefore, the moles of MnO₂ deposited is: mol MnO₂ = 224.01 mol e- / 2 mol e-/mol MnO₂ = 112.0 mol MnO₂ Finally, we can calculate the mass of MnO₂ using its molar mass: mass MnO₂ = mol MnO₂ * molar mass MnO₂ mass MnO₂ = 112.0 mol * 865.19 g/mol = 96,991 g = 9.73 kg Therefore, the answer is (C) 9.73 kg.

A Current of 250