A solution of is electrolytically reduced to Mn³⁺.A current of 5.19 A is passed through the solution for 15.0 min.What is the number of moles of Mn³⁺ produced in this process? (1 faraday = 96,486 coulombs)

Question

Quizplus · Accepted Answer

First, we need to balance the given equation:

11 ea7a3a9a_9f48_c708_a82d_d5b1b37887d5_TB2288_11 + 32 e- → MnS1U1P13 + 16 H+

The number of moles of MnS1U1P13 produced can be calculated using Faraday's law:

moles of MnS1U1P13 = (current x time) / (Faraday's constant x number of electrons transferred in the balanced equation)

number of electrons transferred in the balanced equation = 32

Faraday's constant = 96,486 C

current = 5.19 A

time = 15.0 min = 900 s

moles of MnS1U1P13 = (5.19 A x 900 s) / (96,486 C/mol x 32) = 0.0161 mol

Therefore, the answer is C: 0.0161 mol.

A Solution of Is Electrolytically Reduced to Mn3+

A Solution of Is Electrolytically Reduced to Mn³⁺