<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation
You are viewing an older version of this Concept. Go to the latest version.

Half-Reaction Method in Basic Solution

Explains how to balance redox half-reactions in basic solutions.

Atoms Practice
Practice Half-Reaction Method in Basic Solution
Practice Now
Half-Reaction Method in Basic Solution

Credit: Courtesy of Tomas Sennett, Environmental Protection Agency
Source: http://commons.wikimedia.org/wiki/File:WATER_TREATMENT_PLANT_AT_SUNSET_-_NARA_-_542979.jpg
License: CC BY-NC 3.0

How do you remove cyanide from water ?

Cyanide is a very toxic material. Generated mainly by industrial manufacturing processes, this anion can cause neurological effects and damage to sensitive tissues such as the thyroid gland.

Treatment with chlorine gas in basic solution effectively destroys any cyanide present by converting it to harmless nitrogen gas. The reaction is as follows:

2 \text{NaCN}+5 \text{Cl}_2 +12 \text{NaOH} \rightarrow \text{N}_2 + 2\text{Na}_2\text{CO}_3 + 10\text{NaCl} + 6\text{H}_2\text{O}

Half-Reaction Method in Basic Solution

For reactions that occur in basic solution rather than acidic solution, the steps to balance the reaction are primarily the same. However, after finishing step 6, add an equal number of OH ions to both sides of the equation. Combine the H + and OH to make H 2 O and cancel out any water molecules that appear on both sides. Using the example of the oxidation of Fe 2+ ions by dichromate (Cr 2 O 7 2- ), we would get the following three steps:

1. Adding the hydroxide ions

14\text{OH}^-(aq)+14\text{H}^+(aq)+6\text{Fe}^{2+}(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow 6\text{Fe}^{3+}+2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)+14\text{OH}^-(aq)

2. Combining the hydrogen ions and hydroxide ions to make water

14\text{H}_2\text{O}(l)+6\text{Fe}^{2+}(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow 6\text{Fe}^{3+}+2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)+14\text{OH}^-(aq)

3. Canceling out seven water molecules from both sides to get the final equation

7\text{H}_2\text{O}(l)+6\text{Fe}^{2+}(aq)+\text{Cr}_2\text{O}^{2-}_7(aq) \rightarrow 6\text{Fe}^{3+}+2\text{Cr}^{3+}(aq)+14\text{OH}^-(aq)

The equation is still balanced by atoms and by charge, but the presence of hydroxide ions rather than hydrogen ions means that the reaction takes place in basic solution. Typically, most redox reactions will actually only proceed in one type of solution or the other. The oxidation of Fe 2+ by Cr 2 O 7 2− does not occur in basic solution, and was only balanced this way to demonstrate the method.

In summary, the choice of which balancing method to use depends on the kind of reaction. The oxidation-number method works best if the oxidized and reduced species appear only once on each side of the equation and if no acids or bases are present. The half-reaction method is more versatile and works well for reactions involving ions in aqueous solution.


  • The half-reaction method of balancing redox equations in basic solution is described.


Balance the equations at the link below:



  1. How many OH - should be added to each side?
  2. When does the oxidation-number approach work best?
  3. When does the half-reaction method work best?

Image Attributions

  1. [1]^ Credit: Courtesy of Tomas Sennett, Environmental Protection Agency; Source: http://commons.wikimedia.org/wiki/File:WATER_TREATMENT_PLANT_AT_SUNSET_-_NARA_-_542979.jpg; License: CC BY-NC 3.0

Explore More

Sign in to explore more, including practice questions and solutions for Half-Reaction Method in Basic Solution.


Please wait...
Please wait...

Original text