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Electrochemistry
Question
Consider the reaction: Cr2O72– + 14H+ + 6e- → 2Cr3+ + 8H2O
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72-?
Solution
Answer:
Cr2O72– → 2Cr3+
2Cr6+ + 6e– → 2Cr3+
Cr2O72– → 2Cr3+
2Cr6+ + 6e– → 2Cr3+
In given equation there are 6 electrons are required so that n = 6
Use the formula
Required charge = nF
Plug the values in this formula we get
Required charge = 6 × 96487 Coulombs
= 578922 Coulombs
= 5.79 × 105 Coulombs
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