ElectrochemistryHard
Question
For reducing one mole of Cr2O72- to Cr3+ the cahrge required is :
Options
A.3 × 96500 coulomb
B.6 × 96500 coulomb
C.0.3 Faradays
D.0.6 Faradays
Solution
Cr2O72+ + 14 H+ + 6e- → 2Cr3+ + 7H2O
For reducing one mole of Cr2O72- charge required = 6 × 96500 coulomb.
For reducing one mole of Cr2O72- charge required = 6 × 96500 coulomb.
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