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Consider the following reaction: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$, $E^0 = 1.51$ V The quantity of electricity required in Faraday to reduce five moles of $\text{MnO}_4^-$ is:
1 mole of $\text{MnO}_4^-$ requires 5 Faraday charge. 5 moles require 25 Faraday.
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