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How many faradays are required to reduce all the chromium in 0.15 Lit of $0 .4 MCr {Cr}_{2}^{- 2}$ to ${Cr}^{+ 3}$ ?

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0.9 F

0.36 F

2.4 F

3.6 F

answer is B.

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Detailed Solution

One mole of ${Cr}_{2} O_{7}^{- 2} = 6 F n = M \times V_{lit} = 0 .15 \times 0 .4 = 0 .06 moles$
$0 .06 moles of {Cr}_{2} O_{7}^{- 2} = ? 0 .06 \times 6 = 0 .36 F$

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