Transcript Analyzing Redox Equations

Balancing Redox Equations
0
0
+2 -2
Analyze Mg + S  MgS
•
•
1.
2.
What is oxidized?
What is reduced?
Assign Oxidation Numbers.
Figure out change in oxidation numbers.
Mg goes from 0 to +2: Oxidation
S goes from 0 to -2: Reduction
2 electrons
Mg + S  MgS
2 electrons
3. Identify what species is oxidized & what
species is reduced.
4. Figure out the 2 half-reactions.
Half-Reactions
0
0
+2 -2
Mg + S  MgS
Mg is oxidized:
S is reduced:
Mg  Mg+2 + 2e-
S + 2e-  S-2
5. Adjust half-reactions so electrons lost
= electrons gained.
6. Add half-reactions.
Mg  Mg+2 + 2e-  S-2
S
+
2e
__________________________________
Mg + S + 2e-  Mg+2 +2e- + S-2
7. Balance everything else by counting
atoms.
0
+1 -1
0
+2 -1
Zn + HCl  H2 + ZnCl2
• Zn goes from 0 to +2: oxidation.
• H goes from +1 to 0: reduction.
• Cl goes from -1 to -1: No change.
2 electrons
Zn + HCl  H2 + ZnCl2
1 electron per H
Zn  Zn+2 + 2e+1 + 2e-  H
2H
2
______________________________________
Zn +
+1
2H
+2e

+2
Zn
+2e
+ H2
Transfer the coefficients!
Zn + 2H+1  H2 + Zn+2
• This is what you’ve got from adding the 2
half-reactions.
Zn + HCl  H2 + ZnCl2
• This is the skeleton equation you started
with.
• Transfer the coefficients.
Zn + 2HCl  H2 + ZnCl2
Balancing Redox Equations
1. Assign oxidation numbers to all atoms in equation.
2. See which elements have changes in oxidation
number.
3. Identify atoms that are oxidized & atoms that are
reduced.
4. Write the half-reactions. Diatomics have to be
written as diatomics.
5. Make the number of electrons lost & gained equal in
magnitude by multiplying half-reactions as needed.
6. Add the half-reactions. Transfer coefficients to
skeleton equation.
7. Balance remainder of equation by counting up
atoms.
0
+1 +5 -2
+2 +5 -2
0
Cu + AgNO3  Cu(NO3)2 + Ag
•
•
•
•
Cu goes from 0 to +2: oxidation.
Ag goes from +1 to 0: reduction.
N goes from +5 to +5: No change.
O goes from -2 to -2: No change.
Half-Reactions
Multiply
by 2
Cu  Cu+2 + 2eAg+1 + 1e-  Ag
Cu  Cu+2 + 2e+1 + 2e-  2Ag
2Ag
______________________
Cu + 2Ag+1 + 2e-  2Ag + Cu+2 + 2e-
Transfer Coefficients
• Compare skeleton equation & sum of halfreactions:
Cu + AgNO3  Ag + Cu(NO3)2
Vs.
Cu + 2Ag+1 + 2e-  2Ag + Cu+2 + 2e• Transfer the coefficients!
Cu + 2AgNO3  2Ag + Cu(NO3)2)
Exception: Do NOT insert the coefficient of any item that
appears in more than one place in the equation.
0
+1 +5 -2
+2 +5 -2
+4 -2
+1 -2
Cu + HNO3  Cu(NO3)2 + NO2 + H2O
1. Assign Oxidation Numbers.
2. Identify which species are oxidized &
which reduced.
Cu from 0 to +2 = oxidized
H from +1 to +1 so no change
O from -2 to -2 so no change
N all starts as +5. Some ends as +5, some
as +4 = reduction
Change of +2
Cu + HNO3  Cu(NO3)2 + NO2 + H2O
2
2
2(Change of -1 ) = -2
3. Find change in oxidation number.
4. Write half-reactions.
Cu  Cu+2 + 2eN+5 + 1e-  N+4
What’s oxidized? What’s reduced?
• What is oxidized?
Cu
• What is reduced?
Can’t just say N.
It’s the N in the HNO3.
Or the HNO3 or N+5.
• What is the oxidizing agent?
• What is the reducing agent?
N+5
Cu
5. Multiply half-reactions as necessary.
Cu  Cu+2 + 2e2N+5 + 2e-  2N+4
• Now the # of electrons lost = # gained.
6. Add half-reactions. Transfer coefficients.
Cu + 2HNO3  Cu(NO3)2 + 2NO2 + H2O
7. Balance remaining atoms by inspection.
Cu + 4HNO3  Cu(NO3)2 + 2NO2 + 2H2O