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Oxidation
- Reduction Reactions
"Redox Reactions" |
| |
Oxidation - Lose electrons
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Oxidation # becomes
more positive
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Reduction - Gain
electrons
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Oxidation # becomes
more negative
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| Determine what gets oxidized, reduced,
the oxidizing agent, & the reducing agent: |
| 1) SnCl2 + PbCl
4 ---> SnCl4 + PbCl
2 |
| 2) 2Sb + 3Cl2
---> 2SbCl3 |
| 3) CuS + H+
+ NO3- ---> Cu+2
+ S + NO + H2O |
| 4) NaI + H2SO
4 ---> H2S + I2
+ Na2SO4 + H2
O |
| 5) SO3 + H2
O ---> H2SO4 |
| Answers: |
|
|
|
Balance Redox Equation using the Half-Reaction Method
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| Step 1:
Write the 2 half-reactions Step 2: Balance atoms if needed Step 3: Balance oxygen with water Step 4: Balance hydrogen with H+ ion Step 5: Balance charge with e - Step 6: Make e- lost = e- gained Step 7: Add & simplify equation |
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Balance:
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NO3
- + I2 ---> IO
3 - + NO2
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Answer:
|
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NO
3- ---> NO2
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I
2 ---> IO3-
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|
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I
2 ---> 2IO3-
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NO
3- ---> NO2
+ H2O
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6H
2O + I2 ---> 2IO3-
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|
2H+ +
NO3- ---> NO2
+ H2O
|
6H
2O +
I2 ---> 2IO3-
+ 12H+
|
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2H+
+ NO3-
+ e-
---> NO2 + H2O
|
6H
2O +
I2 ---> 2IO3-
+ 12H+
+ 10e -
|
|
10 [2H+ + NO
3- + e- ---> NO
2 + H2O]
20H+ + 10NO
3- + 10e- ---> 10NO
2 + 10H2O
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20H+ + 10NO
3- + 10e- +
6H2O +
I2
--->
2IO3-
+ 12H+
+ 10e - + 10NO
2 + 10H2O
|
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8H+
+ 10NO3- + I2 --->
2IO 3- + 10NO2 + 4H2O
|
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Balance:
|
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Mn+2
+ BiO3- ---> Bi+3
+ MnO4-
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Answer:
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BiO3-
---> Bi+3
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Mn+2 ---> MnO
4-
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BiO3-
---> Bi+3 + 3H2
O
|
4H2O + Mn
+2 ---> MnO4-
|
| 6H+ + BiO3
- ---> Bi+3
+ 3H2O |
4H2O + Mn
+2 ---> MnO4-
+ 8H+ |
| 2e- + 6H+ + BiO
3- ---> Bi+3
+ 3H2O |
4H2O + Mn
+2 ---> MnO4-
+ 8H+ + 5e- |
| 5[2e- + 6H+ + BiO
3- ---> Bi+3
+ 3H2O] |
2[4H2O +
Mn+2 ---> MnO4-
+ 8H+ + 5e-] |
| 10e- + 30H+ + 5BiO
3- + 8H
2O + 2Mn+2
---> 5Bi+3 + 15H
2O + 2MnO4-
+ 16H+ + 10e- |
|
14H+
+ 5BiO3- 2Mn+2
---> 5Bi+3
+ 7H2O + 2MnO4-
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