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Unit: Electrochemistry
Half Reactions and Redox
Equations
Half-Reactions
Remember OIL RIG…
Oxidation
Is
Loss of electrons
Reduction
Is
Gain of electrons
Half-Reaction Example
2HCl + Mg  MgCl2 + H2
1. Label all oxidation numbers
2. Determine what is reduced and oxidized
reduction = gain of electrons (more negative #)
oxidation = loss of electrons (more positive #)
Half-Reaction Definitions
2HCl + Mg  MgCl2 + H2
Reducing agent: gives electrons and is OXIDIZED
Oxidizing agent: removes electrons and is REDUCED
Example:
2HCl
H = +1
Cl = -1
+
Mg
Mg = 0

MgCl2
Cl = -1
Mg = +2
+
H2
H=0
The oxidation number of hydrogen becomes more negative,
so hydrogen is reduced. Therefore, hydrogen is the oxidizing
agent (it oxides another element).
The oxidation number of magnesium becomes more positive,
so magnesium is oxidized. Therefore, magnesium is the
reducing agent (it reduces another element).
Balancing Half Reactions in Acidic Solutions
1. Label all oxidation numbers.
2. Determine which element is reduced and which is oxidized.
3. Write the two half reactions (follow solubility rules!) and do
the following to each:






Balance element in question using coefficients
Balance oxygen by adding H2O
Balance hydrogen by adding H+
[leave a blank line here!]
Balance charge by adding eMultiply reactions so that number of e- are equal in both
Balancing Redox Reactions in Acidic Solutions Example #1
H2S + HNO3  H2SO4 + NO2 + H2O
Balancing Redox Reactions in Acidic
Solutions
1. Label all oxidation numbers.
2. Determine which element is reduced and which is oxidized.
3. Write the two half reactions (follow solubility rules!) and do
the following to each:






Balance element in question using coefficients
Balance oxygen by adding H2O
Balance hydrogen by adding H+
[leave a blank line here!]
Balance charge by adding eMultiply reactions so that number of e- are equal in both
4. Add reactions together, recombine ions, cancel things on
both sides, and make the equation look like the original
Balancing Redox Reactions in Acidic Solutions Example #1
H2S + HNO3  H2SO4 + NO2 + H2O
Balancing Redox Reactions in Acidic Solutions Example #2
Cr2O7-2 (aq) + Cl- (aq)  Cr+3 (aq) + Cl2 (g)
H2S + HNO3  H2SO4 + NO2 + H2O
1. Assign oxidation numbers.
2. Write the half-reactions:
- If a compound contains an element that changes oxidation
#, write that compound in ionic form (follow solubility rules)
3. Balance:
- Balance element in question using coefficients
- Balance oxygen by adding H2O
- Balance hydrogen by adding H+ ions (in basic solutions add
OH-)
We can add water or H+ because we are assuming we are
doing this in an acidic, aqueous solution where these
species are readily available
H2S + HNO3  H2SO4 + NO2 + H2O
4. Add electrons to balance the charge
5. Repeat process with the other half-reaction
6. Multiply half reactions so that the number of electrons lost
equals the number gained
7. Add the half reactions
8. Re-combine the ions to form the compounds found in the
original formula.
Balancing Redox Equations – Practice #1
Balance the redox equation below:
NaBr(aq) + Cl2(g)→ NaCl(aq) + Br2(l)
Balancing Redox Equations – Practice #2
Balance the redox equation below:
Fe2O3(s) + CO(g) → Fe(s) + CO2(g)
Balancing Redox Equations – Practice #3
Balance the redox equation below:
I2(s) + OCl-(aq) → IO3- (aq) + Cl-(aq)
Balancing Redox Equations – Practice #4
Balance the redox equation below:
HCOOH(aq) + MnO4-(aq) → CO2 (g) + Mn2+(aq)
Note: HCOOH is a weak acid.
Balancing Redox Reactions in Basic
Solutions
1. Label all oxidation numbers.
2. Determine which element is reduced and which is oxidized.
3. Write the two half reactions (follow solubility rules!) and do
the following to each:






Balance element in question using coefficients
Balance oxygen by adding H2O
Balance hydrogen by adding H+
Add OH- to neutralize H+
Balance charge by adding eMultiply reactions so that number of e- are equal in both
4. Add reactions together, recombine ions, cancel things on
both sides, and make the equation look like the original
Balancing Redox Reactions in Basic Solutions Example #1
Cr+3 (aq) + O2 (g)  Cr2O7-2 (aq) + H2O2 (l)