Balance?
Cr2O7^2- + C2H4O yields C2H4O2 + Cr^3+
(When I say ^ a number followed by a plus or minus that's the charge)
Half-reaction method needed.
To balance the given equation using the half-reaction method, follow these steps:
Step 1: Split the reaction into two half-reactions: oxidation and reduction.
The given equation is:
Cr2O7^2- + C2H4O → C2H4O2 + Cr^3+
Oxidation half-reaction: Cr2O7^2- → Cr^3+
Reduction half-reaction: C2H4O → C2H4O2
Step 2: Balance the elements other than hydrogen and oxygen in each half-reaction.
In the oxidation half-reaction, we have one Cr atom on the left side and one Cr atom on the right side. So, no adjustment is needed for Cr.
In the reduction half-reaction, we have two C atoms on the left side and two C atoms on the right side. So, no adjustment is needed for C.
Step 3: Balance the oxygen atoms by adding water (H2O) molecules to the appropriate side of each half-reaction.
In the oxidation half-reaction, we have seven O atoms on the left side and zero O atoms on the right side. So, we need to add seven H2O molecules to the right side to balance the oxygen atoms.
Cr2O7^2- → Cr^3+ + 7H2O
In the reduction half-reaction, we have one O atom on the left side and two O atoms on the right side. So, we need to add one H2O molecule to the left side to balance the oxygen atoms.
C2H4O + H2O → C2H4O2
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+) to the appropriate side of each half-reaction.
In the oxidation half-reaction, we have zero H atoms on the left side and fourteen H atoms on the right side (7H2O). So, we need to add fourteen H+ ions to the left side to balance the hydrogen atoms.
14H+ + Cr2O7^2- → Cr^3+ + 7H2O
In the reduction half-reaction, we have four H atoms on the left side and six H atoms on the right side (C2H4O2). So, we need to add two H+ ions to the right side to balance the hydrogen atoms.
C2H4O + H2O → C2H4O2 + 2H+
Step 5: Balance the charges by adding electrons (e-) to the appropriate side of each half-reaction.
In the oxidation half-reaction, we have a total charge of -2 on the left side and a total charge of +3 on the right side (Cr^3+). So, we need to add 5 electrons to the left side to balance the charges.
5e- + 14H+ + Cr2O7^2- → Cr^3+ + 7H2O
In the reduction half-reaction, we have a total charge of +2 on the left side (C2H4O) and a total charge of +2 on the right side (C2H4O2 + 2H+). So, no adjustment is needed for charges.
C2H4O + H2O → C2H4O2 + 2H+ + 2e-
Step 6: Multiply the half-reactions as necessary to balance the number of electrons in both half-reactions.
Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 5 to equalize the number of electrons.
10e- + 28H+ + 2Cr2O7^2- → 2Cr^3+ + 14H2O
5C2H4O + 5H2O → 5C2H4O2 + 10H+ + 10e-
Step 7: Combine the two balanced half-reactions into a balanced chemical equation.
Combine the oxidation and reduction half-reactions, canceling out the electrons (e-) from both sides.
10e- + 28H+ + 2Cr2O7^2- + 5C2H4O + 5H2O → 5C2H4O2 + 10H+ + 10e- + 2Cr^3+ + 14H2O
Cancel out the common terms:
28H+ + 2Cr2O7^2- + 5C2H4O + 5H2O → 5C2H4O2 + 2Cr^3+ + 14H2O
Simplify the equation, removing the excess water molecules:
28H+ + 2Cr2O7^2- + 5C2H4O → 5C2H4O2 + 2Cr^3+
Finally, the balanced equation is:
2Cr2O7^2- + 5C2H4O + 28H+ → 5C2H4O2 + 2Cr^3+