Balance the following redox reaction in acidic conditions
Cr2O7
2-
(aq) + HNO2 (aq) → Cr3+(aq) + NO3
-
(aq)
Cr2O7^2-(aq) + HNO2(aq) → Cr^3+(aq) + NO3^-(aq)
Two half equations.
Cr2O7^2- + + 14H^+ + 6e ==> 2Cr^3+ + 7H2O
HNO2 + H2O ==> NO3^- + 2e + 3H^+
Multiply equation 1 by 1 and equation by 3 (to make the electrons equal), then add the two half cells. There will be some items that appear on both sides (H^+ will) so add or subtract from both sides to make H^+ appear on just one side. Post your completed work if you want me to check it.
Best
Sure, here's a balanced equation for you:
2 Cr2O7^2- + 10 HNO2 + 16 H+ → 4 Cr3+ + 10 NO3- + 8 H2O
Remember, if you have any oxidation or reduction, just call 1-800-BALANCE to talk to an equation therapist. They'll help you balance things out!
To balance the given redox reaction in acidic conditions, follow these steps:
Step 1: Split the reaction into two half-reactions: oxidation and reduction.
Oxidation half-reaction:
Cr2O7
2-
(aq) → Cr3+(aq)
Reduction half-reaction:
HNO2 (aq) → NO3
-
(aq)
Step 2: Balance the elements other than hydrogen (H) and oxygen (O) in each half-reaction.
Oxidation half-reaction:
Cr2O7
2-
(aq) → 2 Cr3+(aq)
Reduction half-reaction:
HNO2 (aq) → HNO3(aq)
Step 3: Balance the oxygen (O) atoms by adding water (H2O) to the side that has fewer oxygen atoms.
Oxidation half-reaction:
Cr2O7
2-
(aq) + 14 H+(aq) → 2 Cr3+(aq) + 7 H2O(l)
Reduction half-reaction:
HNO2 (aq) → HNO3(aq) + H2O(l)
Step 4: Balance the hydrogen (H) atoms by adding hydrogen ions (H+) to the side that has fewer hydrogen atoms.
Oxidation half-reaction:
Cr2O7
2-
(aq) + 14 H+(aq) → 2 Cr3+(aq) + 7 H2O(l)
Reduction half-reaction:
HNO2 (aq) + 2 H+(aq) → HNO3(aq) + H2O(l)
Step 5: Balance the charges by adding electrons (e-) to the side that has a higher positive charge.
Oxidation half-reaction:
Cr2O7
2-
(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l)
Reduction half-reaction:
HNO2 (aq) + 2 H+(aq) + 2 e- → HNO3(aq) + H2O(l)
Step 6: Equalize the number of electrons in both half-reactions and combine them.
Multiply the oxidation half-reaction by 2:
2 Cr2O7
2-
(aq) + 28 H+(aq) + 12 e- → 4 Cr3+(aq) + 14 H2O(l)
Multiply the reduction half-reaction by 6:
6 HNO2 (aq) + 12 H+(aq) + 12 e- → 6 HNO3(aq) + 6 H2O(l)
Combine the two half-reactions:
2 Cr2O7
2-
(aq) + 6 HNO2 (aq) + 28 H+(aq) → 4 Cr3+(aq) + 6 HNO3(aq) + 14 H2O(l)
Step 7: Verify that the charges and elements are balanced. In this case, the reaction is balanced, and each side has a total charge of 0.
The balanced equation for the redox reaction in acidic conditions is:
2 Cr2O7
2-
(aq) + 6 HNO2 (aq) + 28 H+(aq) → 4 Cr3+(aq) + 6 HNO3(aq) + 14 H2O(l)
To balance the redox reaction in acidic conditions, we need to follow these steps:
Step 1: Determine the oxidation states of all elements in the reaction.
In this reaction, we can determine the oxidation states as follows:
- Oxygen (O) typically has an oxidation state of -2.
- Hydrogen (H) typically has an oxidation state of +1.
- Chromium (Cr) can have various oxidation states.
Step 2: Write separate half-reactions for the oxidation and reduction processes.
The half-reactions for the given reaction are as follows:
Oxidation half-reaction: Cr2O7
2-
(aq) → Cr3+(aq)
Reduction half-reaction: HNO2(aq) → NO3
-
(aq)
Step 3: Balance all elements except hydrogen and oxygen in both half-reactions.
First, let's balance the elements in the above half-reactions:
Oxidation half-reaction: Cr2O7
2-
(aq) → 2Cr3+(aq)
Reduction half-reaction: HNO2(aq) → NO3
-
(aq) + 2H+
Step 4: Balance the oxygen atoms by adding water (H2O) molecules to the side with the deficit.
In the oxidation half-reaction, there are 7 oxygen atoms on the left side (from Cr2O7
2-) and only 3 oxygen atoms on the right side (from Cr3+). Hence, we need to add 4 water molecules to the right side:
Cr2O7
2-
(aq) → 2Cr3+(aq) + 4H2O(l)
In the reduction half-reaction, there are 2 oxygen atoms on the right side (from NO3
-) and none on the left side (from HNO2). We need to add 2 water molecules to the left side:
HNO2(aq) + 2H2O(l) → NO3
-
(aq) + 2H+
Step 5: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side with the deficit.
In the oxidation half-reaction, there are 14 hydrogen atoms on the left side (from Cr2O7
2-) and 8 hydrogen atoms on the right side (from 4H2O). We need to add 6 hydrogen ions (H+) to the right side:
Cr2O7
2-
(aq) + 14H+ → 2Cr3+(aq) + 4H2O(l)
In the reduction half-reaction, there are 4 hydrogen atoms on the left side (from HNO2) and 2 hydrogen atoms on the right side (from 2H2O). We need to add 2 hydrogen ions (H+) to the right side:
HNO2(aq) + 2H2O(l) + 2H+ → NO3
-
(aq) + 2H+
Step 6: Balance the charges by adding electrons (e-) to the side with the charge deficit.
In the oxidation half-reaction, the charge on the left side is -2 (from Cr2O7
2-) and the charge on the right side is +6 (from 2Cr3+). We need to add 8 electrons (e-) to the left side:
Cr2O7
2-
(aq) + 14H+ + 6e- → 2Cr3+(aq) + 4H2O(l)
In the reduction half-reaction, the charge on the left side is +1 (from HNO2) and the charge on the right side is +1 (from NO3
-). Since the charges are already balanced, we don't need to add any electrons.
Step 7: Multiply the half-reactions by appropriate integers to ensure that the number of electrons transferred is the same in both half-reactions.
To make the number of electrons equal in both half-reactions, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 6:
3(Cr2O7
2-
(aq) + 14H+ + 6e-) → 6(Cr3+(aq) + 4H2O(l))
6(HNO2(aq) + 2H2O(l) + 2H+ → NO3
-
(aq) + 2H+)
Step 8: Add the balanced half-reactions together and cancel out any common terms.
Now, we can add the balanced half-reactions together:
3Cr2O7
2-
(aq) + 14H+ + 6HNO2(aq) + 12H2O(l) + 12H+ → 6Cr3+(aq) + 24H2O(l) + 6NO3
-
(aq) + 12H+
Simplifying the equation, we have:
3Cr2O7
2-
(aq) + 6HNO2(aq) + 6H+ → 6Cr3+(aq) + 6NO3
-
(aq) + 12H2O(l)
Therefore, the balanced redox reaction in acidic conditions is:
3Cr2O7
2-
(aq) + 6HNO2(aq) + 6H+ → 6Cr3+(aq) + 6NO3
-
(aq) + 12H2O(l)