1.

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-) ?

Answer»

Solution :* 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


Discussion

No Comment Found