<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Half-Reaction Method in Basic Solution

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

Estimated10 minsto complete
%
Progress
Practice Half-Reaction Method in Basic Solution

MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated10 minsto complete
%
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

#### Howdo 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:

\begin{align*}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}\end{align*}

### 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 H2O and cancel out any water molecules that appear on both sides. Using the example of the oxidation of Fe2+ ions by dichromate (Cr2O72-), we would get the following three steps:

\begin{align*}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)\end{align*}

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

\begin{align*}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)\end{align*}

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

\begin{align*}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)\end{align*}

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 Fe2+ by Cr2O72− 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.

### Review

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?

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes