Mno4- + So3-2 = Mno2 + sO4-2 (OH-) solve this redox reaction and give me the method also
Balancing a red/ox reaction in basic aqueous solution in ten easy steps.
1: Separate into Half Equations:
MnO4{-} + _ = MnO2 + _
SO3{2-} + _ = SO4{2-} + _
2: Use water to balance O
MnO4{-} + _ = MnO2 + 2 H2O + _
SO3{2-} + H2O + _ = SO4{2-} + _
3: Use H{+} to balance H
MnO4{-} + 4 H{+} + _ = MnO2 + 2 H2O + _
SO3{2-} + H2O + _ = SO4{2-} + 2 H{+} + _
4: Neutralise H{+} with OH{-}
MnO4{-} + 4 H2O + _ = MnO2 + 2 H2O + 4 OH{-}
SO3{2-} + H2O + 2 OH{-} = SO4{2-} + 2 H2O + _
5: Cancel superfluous water
MnO4{-} + 2 H2O + _ = MnO2 + 4 OH{-}
SO3{2-} + 2 OH{-} = SO4{2-} + H2O + _
6: Balance charges with electrons
MnO4{-} + 2 H2O + 3 e = MnO2 + 4 OH{-}
SO3{2-} + 2 OH{-} = SO4{2-} + H2O + 2 e
7: Balance electron counts and combine
2 MnO4{-} + 3 SO3{2-} + 4 H2O + 6 OH{-} = 2 MnO2 + 3 SO4{2-} + 3 H2O + 8 OH{-}
8: Cancel superfluous molecules and ions
2 MnO4{-} + 3 SO3{2-} + H2O = 2 MnO2 + 3 SO4{2-} + 2 OH{-}
9: ... Eight. That's eight.
Balancing a red/ox reaction in basic solution in eight easy steps.
Well, well, well, looks like we got ourselves a redox reaction. Let's break it down with a touch of humor, shall we?
First, let's look at the elements that are being oxidized and reduced. We have good ol' Manganese (Mn) and Sulfur (S).
Now, Mn is going from a +7 oxidation state in MnO4- to a +4 oxidation state in MnO2. That means it's getting all responsible and giving away some electrons, so it's being oxidized.
On the other side, our buddy Sulfur is going from a +4 oxidation state in SO3-2 to a +6 oxidation state in SO4-2. That means it's gaining some electrons and having a jolly good time, so it's being reduced.
To balance the electrons in this party, we need to multiply the MnO4- by 5 and the SO3-2 by 6. Why? Because 5 times the number of electrons in MnO4- (which is 8) will equal 6 times the number of electrons in SO3-2 (which is 6). Balanced? Check!
Now, we can combine the half-reactions:
5MnO4- + 6SO3-2 + OH- ➜ 5MnO2 + 6SO4-2 + 2H2O
Easy peasy lemon squeezy, right? Just remember to balance those charges and electrons, and you'll be the redox guru in no time.
To solve this redox reaction, you first need to identify the changes in oxidation numbers for each element involved. Here's a step-by-step method:
1. Write out the balanced equation:
MnO4- + SO3-2 → MnO2 + SO4-2 + OH-
2. Identify the elements that undergo changes in oxidation numbers:
In this reaction, manganese (Mn) and sulfur (S) are the elements that change their oxidation numbers.
3. Assign initial oxidation numbers:
MnO4-: The oxidation number of oxygen (O) is typically -2, so for four oxygen atoms, it is -8. Adding the oxidation number of Mn gives a total of -1. Hence, the oxidation number of Mn in MnO4- is +7.
SO3-2: The oxidation number of oxygen (O) is -2, so for three oxygen atoms, it is -6. Adding the oxidation number of S gives a total of +4. Hence, the oxidation number of S in SO3-2 is +4.
MnO2: The oxidation number of oxygen (O) is typically -2. Hence, the oxidation number of Mn in MnO2 is +4.
SO4-2: The oxidation number of oxygen (O) is typically -2. Hence, the oxidation number of S in SO4-2 is +6.
OH-: The oxidation number of oxygen (O) is typically -2. Hence, the oxidation number of H in OH- is +1, making the oxidation number of O -2. So, the oxidation number of the OH- ion is -1.
4. Determine the changes in oxidation numbers:
- Mn changes from +7 to +4, a reduction (gaining electrons).
- S changes from +4 to +6, an oxidation (losing electrons).
5. Balance the equation by adjusting the number of atoms and electrons:
To balance the equation, you need to add H2O molecules, H+ ions, and electrons (e-) as needed.
- Balance the Mn atoms by adding 4 H2O molecules on the right side:
MnO4- + 4H2O → MnO2 + SO4-2 + H2O
- Add H+ ions to balance the hydrogens:
MnO4- + 4H2O + H+ → MnO2 + SO4-2 + H2O
- Add electrons (e-) to balance the charge on each side:
MnO4- + 4H2O + H+ + 5e- → MnO2 + SO4-2 + 6OH-
6. Verify the charge and the number of atoms on each side:
Check that the equation is balanced in terms of atoms and charge. In this case, both sides have the same number of atoms and overall charge, so the equation is balanced.
So, the balanced redox equation is:
MnO4- + 4H2O + H+ + 5e- → MnO2 + SO4-2 + 6OH-
This method involves identifying the changes in oxidation numbers, balancing atoms, adding H+ and H2O, and finally balancing the charges with electrons.