Balancing Redox Reactions in Acidic Solutions Learning Goal: To balance a redox

Question

Balancing Redox Reactions in Acidic Solutions

Learning Goal:

To balance a redox equation using the half-reaction method of balancing.

In the half-reaction method of balancing for redox reactions, the overall equation is broken down into two half-reactions. The half-reactions are balanced separately, and then added back together. The general procedure in an acidic solution is given here:

Assign oxidation states to all atoms to identify substances being oxidized and reduced.

Split the reaction into two half-reactions: one for oxidation and one for reduction.

Balance each half-reaction with respect to mass.

Balance all elements except O and H.

Balance O by adding H2O.

Balance H by adding H+.

Balance each half-reaction with respect to charge by adding electrons (e).

If necessary, make the number of electrons in each half-reaction equal by multiplying one or both half-reactions by a small whole number.

Add the half-reactions back together, cancelling out species as necessary.

Use steps 3 and 4 listed in the introduction to rewrite this unbalanced half-reaction Cl2Cl such that it is balanced with respect to mass and charge. Do not include phases in your answer. Express your answer as a chemical equation. Use the symbol e- for an electron.

Explanation / Answer

The reaction I have taken as an example is Cr2O72¯ + SO2 + H+ ---> Cr3+ + HSO4¯ + H2O :

Cr2O72¯ ---> Cr3+
SO2 ---> HSO4¯ These are the two half reactions that are possible.

So if we balnce these two half reactions with no of H+ s and the no of e- ns, we will get the following modified half reactions.

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
2H2O + SO2 ---> HSO4¯ + 3H+ + 2e¯.

Now, Equalizing the no of electrons, we see that

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
6H2O + 3SO2 ---> 3HSO4¯ + 9H+ + 6e¯.

By adding the two above half reactions, The final reaction will be following

5H+ + Cr2O72¯ + 3SO2 ---> 2Cr3+ + 3HSO4¯ + H2O.

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