For reducing one mole of Cr2O2-7 to Cr3+ the charge required is: (a) 3 × 96500 coulomb (b) 6 ×96500 coulomb (c) 0.3 Faradays (d) 0.6 Faradays

(b) 6 ×96500 coulomb Explanation: Cr2O2-7 + 14H+ + 6e- → 2Cr3+ + 7H2O For reducing one mole of Cr2O2-7, charge required = 6 × 96500 coulomb.

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For reducing one mole of Cr2O^2-7 to Cr^3+ the charge required is:

Electrochemistry

Last Post by Freddie12 5 years ago

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03/10/2020 4:38 pm

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Shaneylee18

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For reducing one mole of Cr₂O^2-₇ to Cr³⁺ the charge required is:

(a) 3 × 96500 coulomb

(b) 6 ×96500 coulomb

(d) 0.6 Faradays

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03/10/2020 4:42 pm

Freddie12

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(b) 6 ×96500 coulomb

Explanation:

Cr₂O^2-₇+ 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

For reducing one mole of Cr₂O^2-₇, charge required = 6 × 96500 coulomb.

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