The redox reaction below occurs in acidic solution. Balance it with the smallest whole number coefficients and determine the coefficient for H+.
MnO4-(aq) + HNO2(aq) ---> Mn2+(aq) + NO3-(aq)
Mn goes from an oxidation state of +7 on the left to +2 on the right. N goes from an oxidation state of +3 on the left to +5 on the right.
Step 1: Identify the atoms and their oxidation states in the reactants and products:
Reactants:
Mn: +7 (according to its charge as MnO4-)
O: -2 (according to its charge as MnO4-)
H: +1 (according to its charge as HNO2)
N: +3 (according to its charge as HNO2)
Products:
Mn: +2 (according to its charge as Mn2+)
O: -2 (according to its charge as NO3-)
H: +1 (according to its charge as HNO2)
N: +5 (according to its charge as NO3-)
Step 2: Determine the atoms and the corresponding oxidation states that undergo a change in oxidation state. In this case, the atoms that change oxidation state are Mn and N.
Step 3: Write the reduction half-reaction by balancing the changed atoms and charges:
Reduction half-reaction: MnO4-(aq) + 8H+(aq) + 5e- ---> Mn2+(aq) + 4H2O(l)
Step 4: Write the oxidation half-reaction by balancing the changed atoms and charges:
Oxidation half-reaction: HNO2(aq) ---> NO3-(aq) + 2H+(aq) + 2e-
Step 5: Balance the number of electrons transferred in the half-reactions:
Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to have a total of 10 electrons transferred on each side.
10HNO2(aq) ---> 5NO3-(aq) + 10H+(aq) + 10e-
2MnO4-(aq) + 16H+(aq) + 10e- ---> 2Mn2+(aq) + 8H2O(l)
Step 6: Combine the balanced half-reactions. The electrons must cancel out:
10HNO2(aq) + 2MnO4-(aq) + 16H+(aq) ---> 5NO3-(aq) + 10H+(aq) + 10e- + 2Mn2+(aq) + 8H2O(l)
Step 7: Simplify and cancel out any common terms:
10HNO2(aq) + 2MnO4-(aq) + 16H+(aq) ---> 5NO3-(aq) + 12H2O(l) + 2Mn2+(aq)
The coefficient for H+ is 16.
To balance the given redox reaction, follow these steps:
Step 1: Assign oxidation numbers to each element in the reaction.
- In MnO4-, the oxidation state of Mn is +7, and the oxidation state of O is -2.
- In HNO2, the oxidation states of H and O are +1 and -2, respectively.
- In Mn2+, the oxidation state of Mn is +2.
- In NO3-, the oxidation state of N is +5, and the oxidation state of O is -2.
Step 2: Identify the elements that undergo oxidation and reduction.
- In this reaction, Mn goes from +7 to +2, so it is reduced, and N goes from +3 to +5, so it is oxidized.
Step 3: Write separate half-reactions for oxidation and reduction.
- Half-reaction for oxidation: N goes from +3 to +5, so it loses electrons.
N + 2e- → N5+
- Half-reaction for reduction: Mn goes from +7 to +2, so it gains electrons.
Mn7+ + 5e- → Mn2+
Step 4: Balance the number of atoms and charges in each half-reaction.
- Multiply the oxidation half-reaction by 5 to balance the number of electrons and make it equal to the reduction half-reaction.
5(N + 2e-) → 5(N5+)
- The balanced reduction half-reaction remains as it is.
Step 5: Combine the two half-reactions to cancel out electrons and obtain the overall balanced equation.
5(N + 2e-) + Mn7+ + 5e- → 5(N5+) + Mn2+
5N + MnO4- → 5NO3- + Mn2+
Step 6: Balance the atoms other than H and O.
- Count the number of each type of atom on both sides of the equation.
- In this case, there is one Mn, five N, one Mn, and five O on both sides. They are already balanced.
Step 7: Balance the oxygen atoms by adding water (H2O) molecules to the side deficient in oxygen.
- In this case, there are five O on the reactant side, so add five H2O molecules to the product side.
5N + MnO4- → 5NO3- + Mn2+ + 5H2O
Step 8: Balance the hydrogen atoms by adding H+ ions to the side deficient in hydrogen.
- In this case, there are no H on the reactant side, so add 10 H+ ions to the reactant side.
5N + MnO4- + 10H+ → 5NO3- + Mn2+ + 5H2O
Finally, the balanced equation with the smallest whole number coefficients is:
5N + MnO4- + 10H+ → 5NO3- + Mn2+ + 5H2O
Therefore, the coefficient for H+ in the balanced equation is 10.