How much charge is required for the reduction of \(0.05\) moles of \(MnO_4^-\) to \(MnO_2\)?
Answer & explanation
Correct answer: option 2
The correct answer is option 2. \(3 × 96500 × 0.05 C\).
The reaction can be written as:
\(MnO_4^- + 8H^+ + 5e^- \longrightarrow MnO_2 + 4H_2O\)
The oxidation state of Mn on the left hand side of the equation is +7, and it is +2 on the right hand side.
Thus, it is being reduced by 5 units of electrons to achieve the new oxidation state
So, we see that 5 moles of electrons are required to reduce 1 mole \(MnO_4^-\)
1 mole of electrons carries 3 Faraday, or 96,500 C charge.
So, 0.05 moles of electrons will carry \(3 × 96,500 × 0.05 C\)