Balanced equation for Oxidation-Reduction reaction for Ho2 + Cr(OH)3 = CrO4 + OH. Please show your work
I don't know the identity of HO2. The OH needs a minus sign (OH^-). Cr goes from +3 on the left to +6 on the right. I don't see a reduction but it may be the HO2 (could that be H2O2?). Here is a good site that shows you everything you may need to know about redox.
http://www.chemteam.info/Redox/Redox.html
To balance the oxidation-reduction reaction for Ho2 + Cr(OH)3 = CrO4 + OH, we need to follow a few steps:
Step 1: Identify the oxidation states of each element:
The oxidation state of Ho2 would be +2 since Ho is a lanthanide, and the overall charge on the compound is 0.
The oxidation state of Cr in Cr(OH)3 would be +3, as hydroxide (OH) is a polyatomic ion with a charge of -1.
The oxidation state of Cr in CrO4 would be +6, as the overall charge on the compound is -2.
The oxidation state of O in OH would be -2, as hydroxide (OH) is a polyatomic ion with a charge of -1.
Step 2: Determine the change in oxidation states:
To balance the reaction, we need to determine the change in oxidation states for each element involved. In this case, Cr's oxidation state changes from +3 to +6, resulting in an increase of 3.
Step 3: Balance the atoms other than oxygen and hydrogen:
In this reaction, we have 1 Ho, 1 Cr, and 3 O atoms on the left side. On the right side, we have 1 Cr, 4 O, and 1 O atom in the OH group.
Step 4: Balance the oxygen atoms:
Since there are 3 O atoms on the left side and 4 O atoms on the right side, we need to balance the oxygen by adding H2O to the left side of the equation:
Ho2 + Cr(OH)3 + H2O = CrO4 + OH
Step 5: Balance the hydrogen atoms:
Since there are no hydrogen atoms on the left side and 1 hydrogen atom on the right side (from OH), we need to balance the hydrogen by adding H+ to the left side of the equation:
Ho2 + Cr(OH)3 + H2O + H+ = CrO4 + OH
Step 6: Balance the charges:
Since there is a 2+ charge on the Ho2 compound and a 3+ charge on Cr (III) compound, we need to balance the charges by adding electrons (e-) to the left side of the equation.
Ho2 + Cr(OH)3 + H2O + H+ + 2e- = CrO4 + OH
The balanced equation for the oxidation-reduction reaction is:
2 Ho2 + 3 Cr(OH)3 + 3 H2O + 3 H+ + 6 e- = 3 CrO4 + 6 OH