Complete and balance the following equations using the method of half-reaction. both reactions happen in acidic solutions:
a. Cu(s) + NO3-(aq)--> Cu2+(aq) + NO2(aq)
b. Mn2+(aQ) + NaBiO3(s)--> Bi3+(aq) + MnO4-(aq)
I'm having trouble getting the half-rxn equations and the balancing. it just doesnt click in my head
Dr BOB, can you explain, simply, how you do stage 3. or how you get the number of electrons? I just get myself so confused by it.
9666k
Balancing redox equations in acidic solutions can be quite challenging but with the method of half-reactions, it can become more manageable. Let's go through each equation step-by-step:
a. Cu(s) + NO3-(aq) --> Cu2+(aq) + NO2(aq)
Step 1: Identify which elements are getting oxidized and reduced. In this case, copper (Cu) is getting oxidized from 0 to +2 oxidation state, and nitrogen (N) is getting reduced from +5 to +4.
Step 2: Split the reaction into two half-reactions, one for the oxidation and one for the reduction.
Oxidation half-reaction: Cu(s) --> Cu2+(aq)
Reduction half-reaction: NO3-(aq) --> NO2(aq)
Step 3: Balance the atoms in each half-reaction.
Oxidation half-reaction: Cu(s) --> Cu2+(aq) + 2e-
Reduction half-reaction: 3NO3-(aq) + 2H+(aq) + 2e- --> 3NO2(aq) + H2O(l)
Step 4: Balance the charges by multiplying the half-reactions as required. In this case, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:
3Cu(s) --> 3Cu2+(aq) + 6e-
6NO3-(aq) + 4H+(aq) + 4e- --> 6NO2(aq) + 2H2O(l)
Step 5: Add the balanced half-reactions together and cancel out any common species:
3Cu(s) + 6NO3-(aq) + 4H+(aq) --> 3Cu2+(aq) + 6NO2(aq) + 2H2O(l)
b. Mn2+(aq) + NaBiO3(s) --> Bi3+(aq) + MnO4-(aq)
Step 1: Identify the elements that are getting oxidized and reduced. In this case, manganese (Mn) is getting oxidized from +2 to +7 oxidation state, and bismuth (Bi) is getting reduced from +5 to +3.
Step 2: Split the reaction into two half-reactions.
Oxidation half-reaction: Mn2+(aq) --> MnO4-(aq)
Reduction half-reaction: NaBiO3(s) --> Bi3+(aq)
Step 3: Balance the atoms in each half-reaction.
Oxidation half-reaction: Mn2+(aq) --> MnO4-(aq) + 5e-
Reduction half-reaction: 2NaBiO3(s) + 6H+(aq) + 5e- --> 2Bi3+(aq) + 2Na+(aq) + 3H2O(l)
Step 4: Balance the charges by multiplying the half-reactions as required. Multiply the oxidation half-reaction by 2, and the reduction half-reaction by 5:
2Mn2+(aq) --> 2MnO4-(aq) + 10e-
10NaBiO3(s) + 30H+(aq) + 25e- --> 10Bi3+(aq) + 10Na+(aq) + 15H2O(l)
Step 5: Add the balanced half-reactions together and cancel out any common species:
2Mn2+(aq) + 10NaBiO3(s) + 30H+(aq) --> 2MnO4-(aq) + 10Bi3+(aq) + 10Na+(aq) + 15H2O(l)
Remember, balancing redox equations requires practice and patience. Following the steps provided will help you accurately balance the equations using the method of half-reactions in acidic solutions.
Cu(s) + NO3-(aq)--> Cu2+(aq) + NO2(aq)
Step 1. Separate into two half reactions. Cu ==> Cu^+2
NO3^- ==> NO2
Step 2. Determine oxidation states.
Cu goes from 0 on left to +2 on the right.
Step3. Count the charge and add electrons to balance the charge.
Cu ==> Cu^+2 + 2e
Repeat steps 2 and 3 for NO3^-.
NO3^- ==> NO2
N is +5 on left and 4 on right. Add electrons to account for change in oxidation state.
NO3^- + e ==> NO2
Step 4. Count charge. Add H^+ to balance charge.
-2 on left; 0 on the right.
2H^+ + NO3^- + e ==> NO2
Step 5. Add H2O to balance H^+ added.
2H^+ + NO3^- + e ==> NO2 + H2O
Step 6. Now multiply the NO3^- half cell by 2, the Cu half cell by 1 and add the two equations.
Step 7. Cancel anything common to both sides to arrive at the net ionic equation.
Step 8. If desirable, the net ionic equation can be changed to a molecular equation by adding appropriate ions to either side (but always in equal numbers).