# 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

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1. 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).

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2. 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.

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3. 9666k

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