Question

How much charge in terms of faraday is required for reduction of 1 mol of cr2o72- to cr 3+?

Answer 1

as the charge of cr2o7 is 2 we have to multiply that with 1 faraday that is 96500 so it will be
2×96500=193000

Answer 2

Answer: 8 Faraday

Explanation:  $Cr_2O_7^{2-}+14H^+ +8e^-\rightarrow 2Cr^{2+}+7H_2O$

Chromium in +6 oxidation state in  $Cr_2O_7^{2-}$ 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=$1.6\times 10^{-19}C$

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

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