Balance the following redox reaction and identify the oxidizing and reducing agents
A- Fe2+(aq) + Cr2072-(aq) -->Fe3+(aq) + Cr3+(aq) (acidic)
B- N2H4(g) + ClO3-(aq) -->NO2(g) + Cl- (basic)
Here is a site that shows you how to balance redox equations. Repost with specific questions. I can help you through it but you must learn how to do this.
http://www.chemteam.info/Redox/Redox.html
A- 14H + 6Fe^2+ + Cr2O7^2- --> 6Fe^3+ + 2Cr ^3+ + 7H2O
B- 3N2H4 +4Cl03- --> 6NO2+ 12Cl- + 6H2O
This was all I was able to come up with, am I at lease headed in the right direction?
A is ok. Good work.
B is not right. For example, N and H are OK but I see 4 Cl on the left and 12 on the right. I see 12 O on the left and 18 on the right.
ClO3^- ==> Cl^-
Cl is +5 on the left and -1 on the right, so
ClO3^- + 6e ==> Cl^-
I see -7 charge on the left; -1 on the right so I add 6 OH^- to the right.
ClO3^- + 6e ==> Cl^- + 6OH^-
Now add H2O the left to balance.
ClO3^- + 6e + 3H2O ==> Cl^- + 6OH^-
N2H4 ==> NO2
First we balance the N so we are comparing apples and oranges when we look at electron change.
N2H4 ==> 2NO2
Both N on the left = -4; on the right both N are +8 which is a change of -12e so
N2H4 ==> 2NO2 + 12e
The change on the left is zero and -12 on the right, we add OH to the left
12OH^- + N2H4 ==> 2NO2 + 12e
Add to H2O on the other side to balanced
12OH^- + N2H4 ==> 2NO2 + 12e + 8H2O
Multiply equantion 1 by 2 and equation 2 by 1 and add. Cancel common ions/molecules/charges that appear on both sides. I get this final equation.
2ClO3^- + N2H4 ==> 2Cl^- + 2NO2 + 2H2O
To balance a redox reaction, follow these steps:
Step 1: Assign oxidation numbers to each element in the reaction.
Step 2: Identify the element undergoing oxidation (increasing its oxidation number) and the element undergoing reduction (decreasing its oxidation number).
Step 3: Balance the atoms in the equation by adding coefficients in front of the species.
Step 4: Balance the charges by adding electrons (e-) to the equation.
Step 5: Check that the number of atoms and the charge are balanced.
Now, let's balance the given redox reactions:
A- Fe2+(aq) + Cr2O7^2-(aq) --> Fe3+(aq) + Cr3+(aq) (acidic)
Step 1: Assign oxidation numbers:
Fe2+(aq): +2
Cr2O7^2-(aq): +6 (for Cr), -2 (for each O), total -2 * 7 = -14
Step 2: Identify the oxidizing and reducing agents.
In this case, Fe2+ is oxidized to Fe3+, so Fe2+ is the reducing agent.
Cr2O7^2- is reduced to Cr3+, so Cr2O7^2- is the oxidizing agent.
Step 3: Balance the atoms first:
2Fe2+(aq) + Cr2O7^2-(aq) --> 2Fe3+(aq) + Cr3+(aq)
Step 4: Balance the charges:
Now it is necessary to balance the charges by adding electrons (e-) to the equation. The Cr2O7^2- gains 6 e- to become Cr3+, so we need to add 6 electrons (e-) to the reactants' side:
6e- + 2Fe2+(aq) + Cr2O7^2-(aq) --> 2Fe3+(aq) + Cr3+(aq)
Step 5: Check that the number of atoms and the charge are balanced:
On both sides, there are 2 iron (Fe) atoms, 7 oxygen (O) atoms, 1 chromium (Cr) atom, and the charges are balanced, so the equation is balanced.
B- N2H4(g) + ClO3-(aq) --> NO2(g) + Cl- (basic)
Step 1: Assign oxidation numbers:
N2H4(g): The oxidation number of hydrogen (H) is +1, so the total oxidation number of nitrogen (N) is -2 * 2 + 4 = 0.
ClO3-(aq): The oxidation number of oxygen (O) is -2, so the total oxidation number of chlorine (Cl) is +5.
Step 2: Identify the oxidizing and reducing agents.
In this case, N2H4 is oxidized to NO2, so N2H4 is the reducing agent.
ClO3- is reduced to Cl-, so ClO3- is the oxidizing agent.
Step 3: Balance the atoms:
N2H4(g) + ClO3-(aq) --> 2NO2(g) + Cl-(aq)
Step 4: Balance the charges:
The charges are already balanced, so you do not need to add or remove any electrons in this case.
Step 5: Check that the number of atoms and the charge are balanced:
On both sides, there are 2 nitrogen (N) atoms, 6 hydrogen (H) atoms, 3 oxygen (O) atoms, and the charges are balanced, so the equation is balanced.
Now, you have balanced the given redox reactions and identified the oxidizing and reducing agents in each case.