What can you conclude about the reverse change Cr2O7 to 2 CrO4 and its dependence on hydroxide ions? Produce a balanced equation bu adding the proper number of OH- ions and H2O molecules to the appropriate sides of the equation.
To answer this question, we need to understand the concept of oxidation-reduction (redox) reactions and the reaction between Cr2O7 and CrO4 ions in the presence of hydroxide ions (OH-).
First, let's identify the oxidation states of the chromium atoms in each ion:
- In Cr2O7, each chromium atom has an oxidation state of +6.
- In CrO4, each chromium atom has an oxidation state of +6 as well.
The change from Cr2O7 to CrO4 involves the reduction of Cr(VI) to Cr(VI). This means that there is no change in the oxidation state of the chromium atoms in this reaction.
In a basic solution (alkaline conditions), hydroxide ions (OH-) are present. These OH- ions can react with the Cr2O7 ion to form CrO4 ions. The balanced equation for this reaction is as follows:
Cr2O7^2- + 2OH- → 2CrO4^2- + H2O
In this equation, two hydroxide ions (OH-) are added to each Cr2O7 ion to produce two CrO4 ions. Additionally, one water molecule (H2O) is formed as a product.
To summarize, the reverse change from Cr2O7 to CrO4 in the presence of hydroxide ions (OH-) involves the addition of two OH- ions and results in the formation of two CrO4 ions and one water molecule. The balanced equation for the reaction is Cr2O7^2- + 2OH- → 2CrO4^2- + H2O.