Balance the redox equation by oxidation number method:
Cr2O7 2- + Br- + H+ = Cr 3+ + Br + H2O
Please help!
What's the trouble.
Cr goes from +6 for EACH Cr atom to +3. Br goes from -1 to 0.
Here is a site that tells you how to do it but I can help if you identify the problem.
http://www.chemteam.info/Redox/Redox.html
To balance the given redox equation using the oxidation number method, follow these steps:
Step 1: Assign oxidation numbers to all the elements in the equation.
Cr2O7 2-: The oxidation number of oxygen (O) is usually -2, so the total oxidation number of the seven oxygen atoms is -14. Since the overall charge of the Cr2O7 2- ion is -2, the sum of the oxidation numbers of chromium (Cr) and oxygen should be equal to -2. Therefore, the oxidation number of chromium is +6.
Br-: The oxidation number of bromine (Br) in a compound or ion is -1.
H+: The hydrogen ion (H+) has an oxidation number of +1.
Cr 3+: The oxidation number of chromium (Cr) is +3.
Br: The oxidation number of bromine (Br) in an elemental form is 0.
H2O: The oxidation number of oxygen (O) in water (H2O) is typically -2. Since the hydrogens (H) have an oxidation number of +1, the sum of the oxidation numbers in water is 0.
Step 2: Identify the elements undergoing oxidation and reduction.
In this equation, chromium (Cr) is being reduced from +6 to +3, indicating a reduction reaction. Bromine (Br) is being oxidized from -1 to 0, indicating an oxidation reaction.
Step 3: Balance the atoms other than hydrogen and oxygen.
To equalize the number of Cr atoms on both sides, multiply the Cr 3+ ion on the right side by 2.
Cr2O7 2- + Br- + H+ = 2Cr 3+ + Br + H2O
Step 4: Balance the oxygens by adding water (H2O).
Since there are seven oxygen atoms on the left side (from Cr2O7 2-) and only one oxygen atom on the right side (from H2O), we need to add six water molecules (H2O) on the right side.
Cr2O7 2- + Br- + H+ = 2Cr 3+ + Br + 6H2O
Step 5: Balance the hydrogens by adding H+ ions.
There are seven hydrogen atoms on the right side (from 6H2O and the H+ ion), so we need to add seven H+ ions on the left side.
Cr2O7 2- + Br- + 7H+ = 2Cr 3+ + Br + 6H2O
Step 6: Balance the charges by adding electrons.
To neutralize the overall charge of the left side (-2 from Cr2O7 2- and -1 from Br-), we need to add three electrons (e-) to the left side.
Cr2O7 2- + 6e- + 14H+ + Br- = 2Cr 3+ + Br + 6H2O
This is the balanced redox equation using the oxidation number method:
Cr2O7 2- + 6e- + 14H+ + Br- = 2Cr 3+ + Br + 6H2O