Question

cr2O72- + c2o42- balance the chemical equation ​

Answer 1

Answer:

I hope this things helps U a lot..

Answer 2

Answer:

The balanced equation is :

$6CO_2+ 2Cr^{3+}+7H_2O$ →  $Cr_2O_7^{2-} + C_2O_4^{2-}$ → $Cr^{3+} + CO_2$

Explanation:

In the reactant side oxidation number of $Cr$ in $Cr_2O_7^{2-}$ is $+6$ and $C$ in  $C_2O_4^{2-}$ is $+3$.

In the product side oxidation number of $Cr$ in  $Cr^{3+}$ is $+3$ and $C$ in $CO_2$ is $+4$.

Now,

$C_2O_4^{2-}$ → $2CO_2+ 2e^-$       Let this be (i)

$Cr_2O_7^{2-} + 14H^++ 6e^-$ → $2Cr^{3+} + 7H_2O$     Let this be (ii)

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

We add $e^-$ to the side where the charge is excess, this means we will add $6e^-$ in reactant side

Now, to balance charge in (i) :

Multiply (i) by $3$

$3C_2O_4^{2-}$ → $6CO_2 + 6e^-$     Let this be (iii)

From (ii) and (iii) :

$3C_2O_4^{2-} + Cr_2O_7^{2-} + 14H^+$ →  $6CO_2+ 2Cr^{3+}+7H_2O$

The charge $6e^-$ gets cancelled because it is present in both reactant side and product side.

Hence, the balanced equation is :

$6CO_2+ 2Cr^{3+}+7H_2O$ → $6CO_2+ 2Cr^{3+}+7H_2O$$6CO_2+ 2Cr^{3+}+7H_2O$