Question

How many coulombs of electricity are required for reduction of 1 mole of \[\ce{Cr2O^{2-}_{ 7}}\] to Cr³⁺?

Answer 1

To calculate the coulombs of electricity required for the reduction of 1 mole of \[\ce{Cr2O^{2-}_{ 7}}\] to Cr³⁺, follow these steps:

\[\ce{Cr2O^{2-}_{ 7} + 14H+ + 6e− -> 2Cr^{3+} + 7H2O}\]

1 mole of \[\ce{Cr2O^{2-}_{ 7}}\] requires 6 moles of electrons.

By using Faraday’s law

Charge (Q) = n × F    ...(i)

Where:

n = 6 mol e⁻ (from above)

F = 96500 C/mol e⁻

Putting these values in equation (i), we get

Q = 6 × 96500

Q = 579000 C