CBSE Class 11-science Answered

Cr₂O₇^2-(aq) + 3C₂O₄^2-(aq) + 14H⁺(aq)  -->  2Cr³⁺(aq) + 6CO₂(g) + 7H₂O(l)

First Write the skeletons of the oxidation and reduction half-reactions.

Cr₂O₇^2-  -->  Cr³⁺

3C₂O₄^2-(aq) -- > 6CO₂(g)

Step #2:    Balance all elements other than H and O.

To balance the chromium atoms in our first half-reaction, we need a two in front of Cr³⁺.

Cr₂O₇^2-  -->  2Cr³⁺

3C₂O₄^2-(aq) -- > 6CO₂(g)

Step #3:    Balance the oxygen atoms by adding H₂O molecules on the side of the arrow where O atoms are needed.

The first half-reaction needs seven oxygen atoms on the right, so we add seven H₂O molecules.

Cr₂O₇^2-  -->  2Cr³⁺  +  7H₂O

The first half-reaction needs 14 hydrogen atoms on the left to balance the 14 hydrogen atoms in the 7 H₂O molecules, so we add 14 H⁺ ions to the left. 

Cr₂O₇^2-  +  14H⁺  -->  2Cr³⁺  +  7H₂O

The second half-reaction has oxygen which is balanced.

3C₂O₄^2-(aq) -- > 6CO₂(g)

Step #5:    Balance the charge by adding electrons, e-.

The sum of the charges on the left side of the chromium half-reaction is +12 (-2 for the Cr₂O₇^2- plus +14 for the 14 H⁺). The sum of the charges on the right side of the chromium half-reaction is +6 (for the 2 Cr³⁺). If we add six electrons to the left side, the sum of the charges on each side of the equation becomes +6.

6e^-  +  Cr₂O₇^2-  +  14H⁺  -->  2Cr³⁺  +  7H₂O

Similarly, for the other half reaction:                    

3C₂O₄^2-(aq) -- > 6CO₂(g) + 6e^-

Add the 2 half-reactions as if they were mathematical equations.

6e^-  +  Cr₂O₇^2-  +  14H⁺  -->  2Cr³⁺  +  7H₂O

3C₂O₄^2-(aq) -- > 6CO₂(g) + 6e^-

Cr₂O₇^2-(aq) + 3C₂O₄^2-(aq) + 14H⁺(aq)  -->  2Cr³⁺(aq) + 6CO₂(g) + 7H₂O(l)