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Question
Consider the reaction: \[\ce{Cr2O^{2-}_7 + 14H^+ + 6e^- -> 2Cr^{3+} + 7H2O}\]
What is the quantity of electricity in coulombs needed to reduce 1 mol of \[\ce{Cr2O^{2-}_7}\]?
Solution
According to the given reaction,
\[\ce{Cr2O^{2-}_7}\] One mole of ions requires 6 moles of electrons.
∴ F = 6 × 96500 C
= 579000 C
∴ 579000 C of electricity will be required for the reduction of Cr3+.
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