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How much charge is required for the reduction of \(0.05\) moles of \(MnO_4^-\) to \(MnO_2\)? |
\(5 × 96500 × 0.05 C\) \(3 × 96500 × 0.05 C\) \(1 × 96500 × 0.05 C\) \(2 × 96500 × 0.05 C\) |
\(3 × 96500 × 0.05 C\) |
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 \(3 × 96,500 × 0.05 C\) |