Question about Redox Balancing Rxn.

Question

Balance the redox reaction.Indicate the reducing agent and the oxidizing agent for each reaction.

Cr₂O₇^2- + U⁴⁺ → Cr³⁺ + UO₂²⁺ (acidic)

Answer 1

To determine what is oxidized and what is reduced, just compare oxidation numbers for each element on both sides of the equation.

-- Cr in Cr₂O₇^2- has o.n. of 6+. Cr on the right side is 3+. Cr has gained electrons so it has been reduced making it the oxidizing agent.

-- U on the left is 4+ and on the right in UO₂²⁺ it is 6+. U has lost electrons so it has been oxidized making it the reducing agent.

Now for balancing. Treat each reaction separately and add them at the end. Here is a step by step...

Cr₂O₇^2- ==> 2Cr³⁺ ... balanced for Cr

Cr₂O₇^2- ==> 2Cr³⁺ + 7H₂O ... balanced for O

Cr₂O₇^2- + 14H⁺ ==> 2Cr³⁺ + 7H₂O ... balanced for H (in acidic medium)

Cr₂O₇^2- + 14H⁺ + 6e- ==> 2Cr³⁺ + 7H₂O ... balanced for charge and is the balanced eq. for reduction rx

U⁴⁺ ==> UO₂²⁺

U⁴⁺ + 2H₂O ==> UO₂²⁺ ... balanced for U and O

U⁴⁺ + 2H₂O ==> UO₂²⁺ + 4H⁺ ... balanced for H (in acidic medium)

U⁴⁺ + 2H₂O ==> UO₂²⁺ + 4H⁺ + 2e- ... balanced for charge and is the balanced eq. for oxidation rx

We need to multiply this oxidation rx by 3 to make the electrons equal to those in the reduction rx.

Then add the two reactions together and combine and/or cancel like items as follows:

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

3U⁴⁺ + 6H₂O ==> 3UO₂²⁺ + 12H⁺ + 6e-

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Cr₂O₇^2- + 14H⁺ + 6e- + 3U⁴⁺ + 6H₂O ==> 2Cr³⁺ + 7H₂O + 3UO₂²⁺ + 12H⁺ + 6e-

Cr₂O₇^2- + 2H⁺ + 3U⁴⁺ ==> 2Cr³⁺ + H₂O + 3UO₂²⁺ ... BALANCED REDOX EQUATION