Cr2O72- + c2o42- balance the chemical equation

1.	Cr2O72- + c2o42- balance the chemical equation
Answer» Explanation: Cr2O7 (2-) + C2O4(2-) ----> Cr(3+) + CO2 In reactant SIDE OXD number of : Cr in Cr2O7(2-) =+6 and C in C2O4(2-) = +3 In product side oxd number of Cr in Cr(3+) = +3 and C in CO2 = +4 Now, C2O4(2-) -----> 2CO2 + 2e- ....... ( Eq 1) Cr2O7(2-) + 14H+ + 6e- -----> 2Cr(3+) + 7H2O..... (Eq2) (total charge in reactant side is +12 and in product side is +6) {We add e- in the side where charge is more..... This means we will add 6e- in reactant side} Now.... To balance charge in( eq 1)....multiply( eq 1) by 3 ==> 3C2O4(2-) -----> 6CO2 + 6e- ...... (Eq 3) From 2 and 3 3C2O4(2-) + Cr2O7(2-) + 14H+ ---> 6CO2 + 2Cr(3+) +7H2O The charge ie 6e- gets cancelled becoz one is on the reactant side and other is on product side I hope it helps you☺️

Answer»

Explanation:

Cr2O7 (2-) + C2O4(2-) ----> Cr(3+) + CO2

In reactant SIDE OXD number of : Cr in Cr2O7(2-) =+6 and C in C2O4(2-) = +3

In product side oxd number of Cr in Cr(3+) = +3 and C in CO2 = +4

Now,

C2O4(2-) -----> 2CO2 + 2e- ....... ( Eq 1)

Cr2O7(2-) + 14H+ + 6e- -----> 2Cr(3+) + 7H2O..... (Eq2)

(total charge in reactant side is +12 and in product side is +6)

{We add e- in the side where charge is more..... This means we will add 6e- in reactant side}

Now.... To balance charge in( eq 1)....multiply( eq 1) by 3

==> 3C2O4(2-) -----> 6CO2 + 6e- ...... (Eq 3)

From 2 and 3

3C2O4(2-) + Cr2O7(2-) + 14H+ ---> 6CO2 + 2Cr(3+)

+7H2O

The charge ie 6e- gets cancelled becoz one is on the reactant side and other is on product side

I hope it helps you☺️

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