Question

How many faradays are needed for reduction of 5mole of cr2o7 into cr+3?

Answer 1

Explanation:

Explanation: Cr_2O_7^{2-}+14H^+ +8e^-\rightarrow 2Cr^{2+}+7H_2OCr

2

O

7

2−

+14H

+

+8e

−

→2Cr

2+

+7H

2

O

Chromium in +6 oxidation state in Cr_2O_7^{2-}Cr

2

O

7

2−

changes to chromium in +2 oxidation state. Thus 4 mole of electrons are required for 1 mole of chromium an 8 mole of electrons for 2 moles of chromium.

1 electron carry charge=1.6\times 10^{-19}C1.6×10

−19

C

1 mole of electrons carry charge=\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C=1 Faraday

1

1.6×10

−19

×6.023×10

23

=96500C=1Faraday

Thus 8 mole of electrons carry charge=\frac{1F}{1}\times 8=8Faraday

1

1F

×8=8Faraday

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