Balance following oxidation reduction reactions using oxidation states method?
a) Cl2(g) + Al(s) --> Al3+(aq) + Cl-
b.) O2(g) + H2O(l) + Pb(s) --> Pb(OH)2(s)
c) H+(aq) + MnO4-(aq) + Fe2+(aq) --> Mn2+(aq) + Fe3+(aq) + H2O(l)
a) Sure, let's balance the reaction!
We start by assigning oxidation numbers to each element in the reaction:
Cl2(g) + Al(s) --> Al3+(aq) + Cl-
Cl: 0
Al: 0
Al3+: +3
Cl-: -1
Now, let's balance the oxidation-reduction half-reactions:
Oxidation half-reaction: Cl2(g) --> 2Cl-
Reduction half-reaction: Al(s) --> Al3+
Using the oxidation states method, we balance the number of atoms in both half-reactions:
Cl2(g) + 2e- --> 2Cl-
Al(s) --> Al3+ + 3e-
Now, we need to make the number of electrons transferred in both half-reactions equal, so we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:
3Cl2(g) + 6e- --> 6Cl-
2Al(s) --> 2Al3+ + 6e-
Finally, we add the half-reactions together and cancel out the electrons:
3Cl2(g) + 2Al(s) --> 2Al3+(aq) + 6Cl-
And there you have it! The balanced equation is 3Cl2(g) + 2Al(s) --> 2Al3+(aq) + 6Cl-. Voila!
b.) Alright, let's balance this one!
Starting with assigning oxidation numbers:
O2(g) + H2O(l) + Pb(s) --> Pb(OH)2(s)
O: 0
H: +1
Pb: 0
Pb(OH)2(s):
Pb: +2
O: -2
H: +1
Now, let's balance the oxidation-reduction half-reactions:
Oxidation half-reaction: H2O(l) --> 2H+ + 2e-
Reduction half-reaction: Pb(s) --> Pb(OH)2(s) + 4e-
Balancing the number of atoms and charge in both half-reactions:
H2O(l) --> 2H+ + 2e-
4Pb(s) --> 4Pb(OH)2(s) + 16e-
Now, to equalize the number of electrons transferred, we multiply the oxidation half-reaction by 8 and the reduction half-reaction by 2:
8H2O(l) --> 16H+ + 16e-
8Pb(s) --> 8Pb(OH)2(s) + 32e-
Combining the half-reactions and canceling out the electrons:
8H2O(l) + 8Pb(s) --> 8H+ + 8Pb(OH)2(s)
And voila! The equation is 8H2O(l) + 8Pb(s) --> 8H+ + 8Pb(OH)2(s). Well balanced!
c) Alright, let's balance this last one!
Starting with assigning oxidation numbers:
H+(aq) + MnO4-(aq) + Fe2+(aq) --> Mn2+(aq) + Fe3+(aq) + H2O(l)
H: +1
Mn: +7
O: -2
Fe: +2
Mn2+(aq): +2
Fe3+(aq): +3
H2O(l):
H: +1
O: -2
Now, let's balance the oxidation-reduction half-reactions:
Oxidation half-reaction: H+(aq) --> H2O(l) + 2e-
Reduction half-reaction: MnO4-(aq) + 8H+ + 5e- --> Mn2+(aq) + 4H2O(l)
Balancing the number of atoms and charge in both half-reactions:
H+(aq) --> H2O(l) + 2e-
MnO4-(aq) + 8H+ + 5e- --> Mn2+(aq) + 4H2O(l)
Now, to equalize the number of electrons transferred, we multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:
5H+(aq) --> 5H2O(l) + 10e-
2MnO4-(aq) + 16H+ + 10e- --> 2Mn2+(aq) + 8H2O(l)
Combining the half-reactions and canceling out the electrons:
5H+(aq) + 2MnO4-(aq) + 16H+ + 10Fe2+(aq) --> 2Mn2+(aq) + 8H2O(l) + 10Fe3+(aq)
And there you have it! The balanced equation is 5H+(aq) + 2MnO4-(aq) + 16H+ + 10Fe2+(aq) --> 2Mn2+(aq) + 8H2O(l) + 10Fe3+(aq). Well balanced!
a) Cl2(g) + Al(s) --> Al3+(aq) + Cl-
Step 1: Assign oxidation numbers to each element in the equation.
Cl2(g): The oxidation number for Cl is 0.
Al(s): The oxidation number for Al is 0.
Al3+(aq): The oxidation number for Al is +3.
Cl-: The oxidation number for Cl is -1.
Step 2: Identify the atoms undergoing oxidation and reduction.
In this case, Al is being oxidized (going from 0 to +3) and Cl is being reduced (going from 0 to -1).
Step 3: Balance the atoms undergoing oxidation and reduction by adding appropriate coefficients.
Cl2(g) + 2Al(s) --> 2Al3+(aq) + 6Cl-(aq)
b) O2(g) + H2O(l) + Pb(s) --> Pb(OH)2(s)
Step 1: Assign oxidation numbers to each element in the equation.
O2(g): The oxidation number for O is 0.
H2O(l): The oxidation number for H is +1 and for O is -2.
Pb(s): The oxidation number for Pb is 0.
Pb(OH)2(s): The oxidation number for Pb is +2, for O is -2, and for H is +1.
Step 2: Identify the atoms undergoing oxidation and reduction.
In this case, Pb is being oxidized (going from 0 to +2) and O in water is being reduced (going from 0 to -2).
Step 3: Balance the atoms undergoing oxidation and reduction by adding appropriate coefficients.
O2(g) + 2H2O(l) + Pb(s) --> Pb(OH)2(s)
c) H+(aq) + MnO4-(aq) + Fe2+(aq) --> Mn2+(aq) + Fe3+(aq) + H2O(l)
Step 1: Assign oxidation numbers to each element in the equation.
H+(aq): The oxidation number for H is +1.
MnO4-(aq): The oxidation number for O is -2, for Mn is +7.
Fe2+(aq): The oxidation number for Fe is +2.
Mn2+(aq): The oxidation number for Mn is +2.
Fe3+(aq): The oxidation number for Fe is +3.
H2O(l): The oxidation number for H is +1 and for O is -2.
Step 2: Identify the atoms undergoing oxidation and reduction.
In this case, Fe is being oxidized (going from +2 to +3) and Mn in permanganate ion is being reduced (going from +7 to +2).
Step 3: Balance the atoms undergoing oxidation and reduction by adding appropriate coefficients.
5H+(aq) + MnO4-(aq) + 8Fe2+(aq) -> Mn2+(aq) + 8Fe3+(aq) + 4H2O(l)
To balance the oxidation-reduction reactions using the oxidation states method, follow these steps:
a) Cl2(g) + Al(s) --> Al3+(aq) + Cl-
Step 1: Write down the balanced molecular equation without any coefficients:
Cl2(g) + Al(s) --> Al3+(aq) + Cl-
Step 2: Determine the oxidation states of all the elements present in the reaction.
For Cl2(g), the oxidation state of Cl is 0 since it is in its elemental form.
For Al(s), the oxidation state of Al is 0 since it is in its elemental form.
For Al3+(aq), the oxidation state of Al is +3.
For Cl-, the oxidation state of Cl is -1.
Step 3: Identify the elements undergoing oxidation and reduction.
In this reaction, Al is being oxidized from an oxidation state of 0 to +3, and Cl is being reduced from an oxidation state of 0 to -1.
Step 4: Balance the number of electrons transferred between the oxidizing and reducing agent.
Since Al is being oxidized, we need to balance the number of electrons lost by Al with the electrons gained by Cl. The electron transfer can be determined by comparing the change in oxidation states of Al: +3 - 0 = +3, and Cl: -1 - 0 = -1.
Thus, 3 electrons are transferred.
Step 5: Write the half-reactions and balance the number of atoms using coefficients if necessary.
Oxidation half-reaction: Al(s) --> Al3+(aq) + 3e-
Reduction half-reaction: Cl2(g) + 2e- --> 2Cl-
Multiplying the oxidation half-reaction by 2 and the reduction half-reaction by 3 to balance the number of electrons transferred, we have:
2Al(s) --> 2Al3+(aq) + 6e-
3Cl2(g) + 6e- --> 6Cl-
Step 6: Combine the balanced half-reactions to obtain the final balanced equation.
3Cl2(g) + 2Al(s) --> 2Al3+(aq) + 6Cl-
b) O2(g) + H2O(l) + Pb(s) --> Pb(OH)2(s)
Step 1: Write down the balanced molecular equation without any coefficients:
O2(g) + H2O(l) + Pb(s) --> Pb(OH)2(s)
Step 2: Determine the oxidation states of all the elements present in the reaction.
For O2(g), the oxidation state of O is 0 since it is in its elemental form.
For H2O(l), the oxidation state of H is +1 and the oxidation state of O is -2.
For Pb(s), the oxidation state of Pb is 0 since it is in its elemental form.
For Pb(OH)2(s), the oxidation state of Pb is +2, the oxidation state of H is +1, and the oxidation state of O is -2.
Step 3: Identify the elements undergoing oxidation and reduction.
In this reaction, Pb is being oxidized from an oxidation state of 0 to +2, and O is being reduced from an oxidation state of 0 to -2.
Step 4: Balance the number of electrons transferred between the oxidizing and reducing agent.
Since Pb is being oxidized, we need to balance the number of electrons lost by Pb with the electrons gained by O. The electron transfer can be determined by comparing the change in oxidation states of Pb: +2 - 0 = +2, and O: -2 - 0 = -2.
Thus, 4 electrons are transferred.
Step 5: Write the half-reactions and balance the number of atoms using coefficients if necessary.
Oxidation half-reaction: Pb(s) --> Pb2+(aq) + 2e-
Reduction half-reaction: O2(g) + 4e- --> 2O2-
Multiplying the oxidation half-reaction by 2 and the reduction half-reaction by 1 to balance the number of electrons transferred, we have:
2Pb(s) --> 2Pb2+(aq) + 4e-
O2(g) + 4e- --> 2O2-
Step 6: Combine the balanced half-reactions to obtain the final balanced equation.
2Pb(s) + O2(g) + 2H2O(l) --> 2Pb(OH)2(s)
c) H+(aq) + MnO4-(aq) + Fe2+(aq) --> Mn2+(aq) + Fe3+(aq) + H2O(l)
Step 1: Write down the balanced molecular equation without any coefficients:
H+(aq) + MnO4-(aq) + Fe2+(aq) --> Mn2+(aq) + Fe3+(aq) + H2O(l)
Step 2: Determine the oxidation states of all the elements present in the reaction.
For H+, the oxidation state of H is +1.
For MnO4-, the oxidation state of Mn is +7 and the oxidation state of O is -2.
For Fe2+, the oxidation state of Fe is +2.
For Mn2+, the oxidation state of Mn is +2.
For Fe3+, the oxidation state of Fe is +3.
For H2O(l), the oxidation state of H is +1 and the oxidation state of O is -2.
Step 3: Identify the elements undergoing oxidation and reduction.
In this reaction, Mn is being reduced from an oxidation state of +7 to +2, and Fe is being oxidized from an oxidation state of +2 to +3.
Step 4: Balance the number of electrons transferred between the oxidizing and reducing agent.
Since Mn is being reduced, we need to balance the number of electrons gained by Mn with the electrons lost by Fe. The electron transfer can be determined by comparing the change in oxidation states of Mn: +7 - +2 = +5, and Fe: +2 - +3 = -1.
Thus, 5 electrons are transferred.
Step 5: Write the half-reactions and balance the number of atoms using coefficients if necessary.
Oxidation half-reaction: Fe2+(aq) --> Fe3+(aq) + 1e-
Reduction half-reaction: MnO4-(aq) + 5e- --> Mn2+(aq)
Multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 1 to balance the number of electrons transferred, we have:
5Fe2+(aq) --> 5Fe3+(aq) + 5e-
MnO4-(aq) + 5e- --> Mn2+(aq)
Step 6: Combine the balanced half-reactions to obtain the final balanced equation.
8H+(aq) + MnO4-(aq) + 5Fe2+(aq) --> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Cl2(g) + Al(s) --> Al3+(aq) + Cl-
Cl2 + 2e --> 2Cl^-
Al ===> Al^3+ + 3e
Multiply equation 1 by 3 and equation 21 by 2 and add. Cancel the electrons.
Here is MnO4^- from part c done this way in detail
MnO4^- ==> Mn^2+
1. The oxidation state Mn on the left is +7; on the right is +2. Add electrons to the appropriate side to balance the change in electrons.
2. MnO4^- + 5e ==> Mn^2+
3. Count the charge on the left and right, then add H^+ (since this is an acid solution but add OH^- if a basic solution) to the appropriate side to balance the charge..
charge on left is -6; on the right is +2 so I add 8H^+ to left.
4. MnO4^- + 5e + 8H^+ ==> Mn^2+
5. Then add H2O (makes no difference in the previous step if you added H^+ or OH^-) to the appropriate side to balance.
MnO4^- + 5e + 8H^+ ==> Mn^2+ + 4H2O
6. Check it.
a. Mn goes from 7 to 2; add 5e to 7 to make 2.
b. count charge on left and right to see if it balances. 2+ on left; 2+ on right
d. Check that atoms balance.
1Mn on left and right; 8H on left and right; 4 O on left and right.
Success.
Here is an example of one (same MnO4^-) using OH but I won't go through the explanation. You can read that step by step from above.
MnO4^- --> MnO2
Mn is 7 on left and 4 on right. Add 3e
MnO4^- + 3e ==> MnO2
Charge on left is -4; on right is 0. Add 4OH^- to right.
MnO4^- + 3e ==> MnO2 + 4OH^-
Then add H2O to the left.]
MnO4^- + 3e + 2H2O ==> MnO2 + 4OH^-
I'll let you check it but I know it's right because I checked it too. ALWAYS check it; that way you will know if you have it right or wrong. If wrong you can redo it and get it right. There is NEVER and excuse for THINKING you have it right and getting it wrong or THINKING it wrong and getting it right. You CAN know it's wrong but not know how to fix it to right. ;-)