Consider the reaction:
`Cr_(2)O_(2)^(2-)+14H^(+)+6e^(-) to 2Cr^(3+)+7H_(2)O`
What is the quantity of electricity in coulombs needed to reduce 1 mol of `Cr_(2)O_(7)^(2-)` ?

Explanation

* Reaction:

* Therefore, to reduce 1 mole of `Cr_(2)O_(7)^(2-),` 6 mole electrons will be used.
So, the required quantity of electricity will be:
So, 6 mole electron=6F electricity is used.
6 F electricity=6`xx96500` Coulomb
=579000 Coulomb