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:

Cr₂O₇^2– + 14H⁺ + 6e^– → 2Cr³⁺ + 8H₂O

What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂O₇^2– ?

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