What is the balance of the following redox reaction
Fe2+(aq)+Cr2O72-(aq) Cr3+(aq)+Fe3+(aq)
I assume you want the balance of the items to add to the left and right.
Add H^+ to the left and H2O to the right.
oh wait
To determine the balance of the redox reaction, we need to assign oxidation numbers to each element in the reaction.
In this reaction, Fe starts with an oxidation number of +2 and ends with an oxidation number of +3. Cr starts with an oxidation number of +6 and ends with an oxidation number of +3. Oxygen typically has an oxidation number of -2, and the sum of all oxidation numbers in a neutral compound is equal to zero.
Let's balance the reaction step-by-step:
1. Assign oxidation numbers to each element:
Fe2+(aq) --> Fe3+(aq): Fe starts with +2 and ends with +3, gaining one electron.
Cr2O72-(aq) --> Cr3+(aq): Cr starts with +6 and ends with +3, gaining three electrons.
2. Write the half-reactions for each half of the reaction:
Oxidation half-reaction: Fe2+(aq) --> Fe3+(aq) + e-
Reduction half-reaction: Cr2O72-(aq) + 6e- --> 2Cr3+(aq)
3. Balance the number of electrons transferred in each half-reaction:
To balance the electrons, we multiply the oxidation half-reaction by 6:
6Fe2+(aq) --> 6Fe3+(aq) + 6e-
Now, the number of electrons transferred in each half-reaction is the same: 6 electrons.
4. Balance the atoms other than oxygen and hydrogen:
To balance the atoms, we add water (H2O) and hydrogen ions (H+) as needed.
Oxidation half-reaction: 6Fe2+(aq) --> 6Fe3+(aq) + 6e-
Reduction half-reaction: Cr2O72-(aq) + 6e- --> 2Cr3+(aq) + 7H2O
5. Balance oxygen atoms by adding water (H2O):
Oxidation half-reaction: 6Fe2+(aq) --> 6Fe3+(aq) + 6e-
Reduction half-reaction: Cr2O72-(aq) + 6e- --> 2Cr3+(aq) + 7H2O + 6e-
6. Balance hydrogen atoms by adding hydrogen ions (H+):
Oxidation half-reaction: 6Fe2+(aq) + 12H+(aq) --> 6Fe3+(aq) + 6e-
Reduction half-reaction: Cr2O72-(aq) + 6e- --> 2Cr3+(aq) + 7H2O + 6e-
7. Cancel out the electrons:
Oxidation half-reaction: 6Fe2+(aq) + 12H+(aq) --> 6Fe3+(aq)
Reduction half-reaction: Cr2O72-(aq) + 6e- --> 2Cr3+(aq) + 7H2O
8. Add the half-reactions together:
6Fe2+(aq) + 12H+(aq) + Cr2O72-(aq) --> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O
Now, the redox reaction is balanced. The balanced equation is:
6Fe2+(aq) + 12H+(aq) + Cr2O72-(aq) --> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O
To determine the balance of a redox reaction, we need to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. Here's how we can balance the given redox reaction between Fe2+(aq) and Cr2O72-(aq):
1. Start by writing the unbalanced equation:
Fe2+(aq) + Cr2O72-(aq) --> Cr3+(aq) + Fe3+(aq)
2. Split the reaction into two half-reactions: oxidation and reduction.
Oxidation half-reaction:
Fe2+(aq) --> Fe3+(aq)
Reduction half-reaction:
Cr2O72-(aq) --> Cr3+(aq)
3. Balance the atoms other than oxygen and hydrogen in each half-reaction. In the oxidation half-reaction, there is only one atom on each side, so it is already balanced. In the reduction half-reaction, we have:
Cr2O72-(aq) --> 2Cr3+(aq)
4. Balance the oxygen atoms by adding water (H2O) molecules to the side that needs them. In the reduction half-reaction, we have seven oxygen atoms on the left side, and we need to balance it to three on the right side. Therefore, we add four water molecules to the right side:
Cr2O72-(aq) + 4H2O(l) --> 2Cr3+(aq)
5. Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs them. In the reduction half-reaction, there are 8 hydrogen atoms on the left side, and we need to balance it on the right side. Therefore, we add 8 hydrogen ions to the left side:
Cr2O72-(aq) + 4H2O(l) + 8H+(aq) --> 2Cr3+(aq)
6. Now, let's determine the number of electrons transferred in each half-reaction. In the oxidation half-reaction, Fe2+ loses one electron to become Fe3+, so the number of electrons lost is 1. In the reduction half-reaction, each Cr3+ ion is formed from one Cr2O72- ion, which requires 6 electrons. Therefore, the number of electrons gained is 6.
7. To balance the electrons, multiply the oxidation half-reaction by 6 and the reduction half-reaction by 1:
6Fe2+(aq) --> 6Fe3+(aq)
Cr2O72-(aq) + 4H2O(l) + 8H+(aq) --> 2Cr3+(aq)
8. Combine the half-reactions and cancel out the common species on both sides:
6Fe2+(aq) + Cr2O72-(aq) + 4H2O(l) + 8H+(aq) --> 6Fe3+(aq) + 2Cr3+(aq)
Now, the redox reaction is successfully balanced.