Friday, May 24, 2013

What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72– given that Consider the reaction: Cr2O72– + 14H+ + 6e– → 2Cr3+ + 8H2O


Q3.12: Consider the reaction:
Cr2O72–+ 14H+ + 6e → 2Cr3+ + 8H2O
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72– ?
Solution:
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

No comments:

Post a Comment