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Cr^3+ + CO2 + H2O-->

# Balance the following redox reaction by ion electron method: Cr2O7^2- + H^+ + C2O4^2- --> Cr^3+ + CO2 + H2O

Asked by shobhit_dc 9th January 2012, 7:28 PM

Cr2O72-(aq) + 3C2O42-(aq) + 14H+(aq)  -->  2Cr3+(aq) + 6CO2(g) + 7H2O(l)

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

Cr2O72-  -->  Cr3+

3C2O42-(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+.

Cr2O72-  -->  2Cr3+

3C2O42-(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.

Cr2O72-  -->  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.

Cr2O72-  +  14H+  -->  2Cr3+  +  7H2O

The second half-reaction has oxygen which is balanced.

3C2O42-(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 Cr2O72- 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-  +  Cr2O72-  +  14H+  -->  2Cr3+  +  7H2O

Similarly, for the other half reaction:

3C2O42-(aq) -- > 6CO2(g) + 6e-

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

6e-  +  Cr2O72-  +  14H+  -->  2Cr3+  +  7H2O

3C2O42-(aq) -- > 6CO2(g) + 6e-

Cr2O72-(aq) + 3C2O42-(aq) + 14H+(aq)  -->  2Cr3+(aq) + 6CO2(g) + 7H2O(l)

Answered by Expert 10th January 2012, 10:57 AM
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