Question

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Balance the following reactions in an acidic solution:

Cr2O7^2-(aq)+HNO2(aq)--->Cr^3+(aq)+ NO3-(aq)

Pb^2(aq)+ IO3-(aq)--->PbO2(aq)+I2(aq)

Answer 1

1. [Cr2O7]^2-(aq) + [HNO2](aq) --> [Cr]³⁺(aq) + [NO3]^-(aq) in acidic solution

The first thing to do is to find the oxidation number of each element, assuming O is -2 and H is +1. It comes to this:

Reactants:
Cr +6
O -2 (note that O is in both reactants)
H +1
N +3

Products:
Cr +3
O -2
N +5

Because oxidation is loss of electrons and reduction is gain of electrons (OIL RIG), Cr is reduced (+6 to +3) and N is oxidized (+3 to +5).

Now, we separate the reaction into two half reactions:

Oxidation:
[HNO2] --> [NO3]-
Reduction:
[Cr2O7]2- --> [Cr]3+
That was the hard part. Now it is a simple matter to balance the halves. First, balance the non-oxygen or hydrogen atoms:

[HNO2] --> [NO3]- (N is balanced)
[Cr2O7]2- --> 2 [Cr]3+ (becomes 2 [Cr]3+)

Then, balance the oxygens by adding water to one side or the other:

[H2O] + [HNO2] --> [NO3]-
[Cr207]2- --> 2[Cr]3+ + 7[H20]

Balance the hydrogen by adding H+ ions to one side or the other:

[H2O] + [HNO2] --> [NO3]- + 3[H]+
14[H]+ + [Cr207]2- --> 2[Cr]3+ + 7[H20]

Then balance the charge by adding e- as needed:

[H2O] + [HNO2] --> [NO3]- + 3[H]+ + 2e-
6e- + 14[H]+ + [Cr207]2- --> 2[Cr]3+ + 7[H20]

Note that the e- should be on opposite sides. Now you have your two balanced half reactions. It is a simple matter to multiply each reaction in order that the coefficient of each e- is their least commom multiple (in this case, 6):

3[H2O] + 3[HNO2] --> 3[NO3]- + 9[H]+ + 6e-
6e- + 14[H]+ + [Cr207]2- --> 2[Cr]3+ + 7[H20]

Then simply add the reactions together and eliminate intermediates (things on both sides, like e-, H+, or H20):

3[HNO2] + 5[H]+ [Cr207]2- --> 3[NO3]- + 2[Cr]3+ + 4[H20]

Don't forget to check to make sure that both charge and number of each element is the same for both sides of the reaction.

If this was taking place in basic solution, you would simply add OH- ions to each side until there were no H+ ions remaining:

3[HNO2] + [H2O] + [Cr2O7]^2- --> 3[NO3]^- + 2[Cr]³⁺ + 5[OH]^-

2. Pb^2+ -----> PbO2

balance the O by adding H2O to the other side
Pb^2+ + 2H2O ---> PbO2

balance the H you added in the H2O by adding H+ to the other side
Pb^2+ + 2H2O ---> PbO2 + 4H+

Convert to basic condions by adding the same umber of OH- as you have H+ to both sides of the arrow
Pb^2+ + 2H2O +4OH- ---> PbO2 + 4H+ + 4OH-

each 1 H+ and OH- will form water
Pb^2+ + 2H2O +4OH- ---> PbO2 + 4H2O

simplify (cancel out H2O that occurs on both sides)
Pb^2+ + 4OH- ---> PbO2 + 2H2O

work out the electrons, add up the total charge on each side of the arrow and then add enouh electrons to the most positve side to balance the charge
LHS = 1 xPb2+ + 4 x OH- = 2-
RHS = neutral
add 2 e to the RHS

Pb^2+ + 4OH- ---> PbO2 + 2H2O + 2e
balanced oxidation half equation


reducton half equation
IO3^- -----> IO^-
IO3^- ------> IO^- + 2H2O
IO3^- + 4H+ ---->IO^- + 2H2O
IO3^- + 4H+ + 4OH- ---> IO^- + 2H2O + 4OH-
IO3^- + 4H2O ----> IO^- + 2H2O + 4OH-
IO3^- + 2H2O ----> IO^- + 4OH-

IO3^- + 2H2O + 4e ---> IO^- + 4OH-
balanced reduction half equation


The oxidation half equation only transferrs 2 elcetrons wheras the reduction half equation transferrs 4. So you need 2 oxidation half equations for every 1 reduction half equation

2 x (Pb^2+ + 4OH- ---> PbO2 + 2H2O + 2e)

2Pb^2+ + 8OH- ---> 2PbO2 + 4H2O + 4e
IO3^- + 2H2O + 4e ---> IO^- + 4OH-
--------------------------------------...
2Pb^2+ +IO3^- + 8OH- + 2H2O +4e ---> 2PbO2 + IO^- + 4H2O + 4OH- + 4e

simplify (cancel out everything that will cancel out)
2Pb²⁺ +IO3^- + 4OH- ---> 2PbO2 + IO^- + 2H2O