Question 12

Consider the reaction:

Cr2 O72- + 14H+ + 6e- → Cr3+ + 8H2

What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2 O72-?

Answer

The given reaction is as follows:

Cr2 O72- + 14H+ + 6e- → Cr3+ + 8H2

Therefore, to reduce 1 mole of Cr2 O72-, the required quantity of electricity will be:

= 6 F

= 6 × 96487 C

= 578922 C

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