The number of moles of electron required to reduce 0.2 mole of Cr_2O_7^2- to Cr^+3

Question

Accepted Answer

Correct answer: A. (a): Cr_2O_7^2-+6e^- ⟶2 Cr^3++6 Change in oxidation state =+3 per chromium atom. 1 mole of Cr_2O_7^2- required =6e^- ∴0.2 moles of Cr required =0.2 6 =1.2 moles of e^-s

$The number of moles of electron required to reduce 0.2 mole of Cr_{2} O_{7}^{2 -} to Cr^{+ 3}$

Solution

More Redox Reactions PYQs