Mno4- + So3-2 = Mno2 + sO4-2 (in neutral solution,but H+ ions will appear in the final question) solve this redox reaction and give me the method also
Thank you
MnO4- + SO3-- >> MnO2 + SO4--
species before after gain
Mn 7 4 3 electr
O -2 -2 0
S 4 6 -2
therefore, we need 2 Mn for each 3 S
2MnO4- + 3SO3-- > 2MnO2 + 3SO4--
To balance the given redox reaction:
Step 1: Assign oxidation numbers to each element in the reaction.
Let's assign oxidation numbers to Mn and O in each compound:
MnO4-: Mn = +7, O = -2
SO3-2: S = +4, O = -2
MnO2: Mn = +4, O = -2
SO4-2: S = +6, O = -2
Step 2: Identify which elements are being reduced and which are being oxidized.
In this case, Mn goes from +7 to +4 (reduction) and S goes from +4 to +6 (oxidation).
Step 3: Write the two half-reactions for oxidation and reduction.
Reduction half-reaction:
MnO4- + 4H+ + 3e- → MnO2 + 2H2O
Oxidation half-reaction:
SO3-2 → SO4-2 + 2e-
Step 4: Balance the atoms in each half-reaction.
Balance the reduction half-reaction:
MnO4- + 8H+ + 5e- → MnO2 + 4H2O
Balance the oxidation half-reaction:
SO3-2 → SO4-2 + 2e-
Step 5: Balance the charge in each half-reaction by adding electrons.
Balance the reduction half-reaction:
MnO4- + 8H+ + 5e- → MnO2 + 4H2O
Balance the oxidation half-reaction:
2SO3-2 → 2SO4-2 + 4e-
Step 6: Balance the number of electrons in both half-reactions.
Multiply the reduction half-reaction by 4 and the oxidation half-reaction by 5:
4MnO4- + 32H+ + 20e- → 4MnO2 + 16H2O
10SO3-2 → 10SO4-2 + 20e-
Step 7: Combine the half-reactions and cancel out the common species.
Multiply the oxidation half-reaction by 2 and add it to the reduction half-reaction:
4MnO4- + 32H+ + 20SO3-2 → 4MnO2 + 16H2O + 10SO4-2
Step 8: Verify that the atoms and charges are balanced on both sides of the equation.
There are 4 Mn, 32 H, 20 S, and 80 O atoms on both sides of the equation, and the charges are balanced as well, so the equation is balanced.
The method used to solve this redox reaction is the half-reaction method, which involves breaking the reaction into two half-reactions (oxidation and reduction), balancing the atoms and charges in each half-reaction individually, then combining the half-reactions to obtain the balanced overall equation.
To solve the given redox reaction, we will use the method of balancing redox reactions known as the half-reaction method.
Step 1: Assign oxidation numbers to all the elements in the reaction.
In this reaction, Mn is being reduced from +7 to +2, and S is being oxidized from +4 to +6.
Step 2: Separate the reaction into two half-reactions: one for oxidation and one for reduction.
Oxidation half-reaction:
S in So3-2 is being oxidized to So4-2.
So3-2 -> So4-2
Reduction half-reaction:
Mn in Mno4- is being reduced to Mno2.
Mno4- -> Mno2
Step 3: Balance the atoms in each half-reaction.
Oxidation half-reaction:
So3-2 -> So4-2
In order to balance the oxygen atoms, we need to add one water molecule to the left side:
So3-2 -> So4-2 + H2O
Next, balance the sulfur atoms by adding two hydrogen ions to the left side:
So3-2 + 2H+ -> So4-2 + H2O
Reduction half-reaction:
Mno4- -> Mno2
To balance the oxygen atoms, we need to add four water molecules to the right side:
Mno4- -> Mno2 + 4H2O
Then, balance the charge by adding eight electrons to the left side:
Mno4- + 8e- -> Mno2 + 4H2O
Step 4: Balance the electrons in both half-reactions.
To balance the number of electrons, we need to multiply the oxidation half-reaction by eight:
8So3-2 + 16H+ -> 8So4-2 + 8H2O
Now, both half-reactions have an equal number of electrons (8e-).
Step 5: Combine the half-reactions.
Add both half-reactions together:
8So3-2 + 16H+ + Mno4- + 8e- -> 8So4-2 + 8H2O + Mno2
Make sure to cancel out any common species on both sides:
8So3-2 + 4Mno4- + 16H+ -> 8So4-2 + 8H2O + 4Mno2
And there you have it! The balanced redox reaction is:
8So3-2 + 4Mno4- + 16H+ -> 8So4-2 + 8H2O + 4Mno2
Note: In the original question, it is mentioned that the reaction occurs in a neutral solution. However, it is also mentioned that H+ ions appear in the final question. This means that the reaction is not actually taking place in a neutral solution, but rather in an acidic solution where H+ ions are present.