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```Topic: Redox – Half reactions
Assign Oxidation number to H, Cl
and O for the following compounds
1. HClO
2. HClO2
3. HClO3
4. HClO4
Now we know how to assign oxidation
number…we can look at redox rxns
• Haber Process
–N2(g) + 3H2(g)  2NH3(g)
• Start by assigning oxidation numbers
0
0
-3 +1
• N2(g) + 3H2(g)  2NH3(g)
• What was oxidized? Reduced?
0
0
-3 +1
N2(g) + 3H2(g)  2NH3(g)
4•
3•
2
1•
0
-1•
-2
-3•
-4•
O • Oxidation
I • Is
L • Losing electrons
N
Began
Ended
0
-3
Gained 3 electrons = Reduced
R • Reduction
I • Is
H 0
+1
G • Gaining electrons
Lost 1 electron = Oxidized
Why use the word “reduced” when
electrons are gained?
Electrons are Negative!
Look how oxidation number changes
Ex: Cl gains an electron → Cl-1
• oxidation # ↓ from 0 to -1; the # was reduced
Half Reactions
• Even though oxidation & reduction reactions
occur together we write separate equations for
each process and include # of e- gained/lost
– known as Half-Reactions
Half Reactions
• Half-reactions must demonstrate:
– conservation of mass & conservation of charge
– # atoms on left must = # atoms on right
• (always balance mass first!!!!)
• Keep 7 diatomics together, everything else
write as single ion)
– total charge on left must = total charge on right
• Add electrons to more positive side
0
0
3H2 
N2 +
-3 +1
2NH3
• Oxidation = electrons lost OIL
(becomes more positive)
–So electrons are on the product
side
But something
•H2  H+1 + 1e- is wrong!
0
=
+1
1 and -1 equals 0
Remember…
• H2  H+1 + 1e-
• Total Charge on left = Total Charge on right
• # atoms on left = # atoms on right
• H2 
• Something is still wrong!
• Charge is off now
• H2  2H+1 + 2e+1
2H +
1e
2 x +1 = +2 and -2 equals 0
0
N2
+
0
3H2 
-3 +1
2NH3
• Reduction = electrons gained RIG
(becomes more negative)
–So electrons are on the reactant
side
• Each N gained 3 electrons,
6 -  2 N-3
–N2 + 3e
Balance mass
But not you have -6
0
You Try:
•
•
•
•
0
+2 -2
Mg + S  MgS
What is oxidized?
What is reduced?
Assign Oxidation Numbers
Figure out change in oxidation numbers
Mg:
0 to +2 = Oxidation
S:
0 to -2 = Reduction
Now write the Half-Reactions
0
0
+2 -2
Mg + S  MgS
Mg is oxidized:
S is reduced:
Mg  Mg+2 + 2e-
S + 2e-  S-2
0
+1 -1
0
+2 -1
Zn + 2HCl  H2 + ZnCl2
What is oxidized? Reduced?
• Zn goes from 0 to +2 = oxidation
• H goes from +1 to 0 = reduction
• Cl goes from -1 to -1; No change
0
+1 -1
0
+2 -1
Zn + 2HCl  H2 + ZnCl2
• Write Half Reactions
Zn  Zn+2 + 2e2H+1 + 2e-  H2
In Redox # of electrons gained has to
equal number of electrons lost
Zn  Zn+2 + 2e2H+1 + 2e-  H2
H2  2H+1 + 2eN2 + 6e-  2N-3
In Redox # of electrons gained has to
equal number of electrons lost
3 H2 
+1
2H +
3H2 
2e
+1
6H +
N2 + 6e-  2N-3
3H2 + N2  6H+1 + 2N-3
6e
N2(g) + 3H2(g)  2NH3(g)
3H2 + N2  6H+1 + 2N-3
1. Assign oxidation numbers to all atoms in equation
2. Determine elements changed oxidation number
3. Identify element oxidized & element reduced
4. Write half-reactions (diatomics must stay as is, everyone
you can write with their oxidation number only
–
NH3 = N-3 but Cl2 stays Cl2 not Cl)
–
Make sure # of atoms on each side is balanced
–
Make sure charge on each side is balanced
5. Number electrons lost & gained must be equal; multiply
half-reactions if necessary
0
+1 +5 -2
+2 +5 -2
0
Cu + 2AgNO3  Cu(NO3)2 + 2Ag
1. Assign oxidation numbers to all atoms in
equation
2. Determine elements changed oxidation
number
– Cu 0+2
Ag +1 0
– N +5  +5 unchanged
O -2  -2 unchanged
3. Identify element oxidized & element reduced
– Oxidation (OIL) =Cu
– Reduction (RIG) = Ag
0
+1 +5 -2
+2 +5 -2
0
Cu + 2AgNO3  Cu(NO3)2 + 2Ag
4. Write half-reactions
– Cu  Cu+2 + 2e– Ag+1 + 1e-  Ag
5. Number electrons lost & gained
must be equal; multiply halfreactions if necessary
– 2(Ag+1 + 1e-  Ag)
– 2Ag+1 + 2e-  2Ag
0
+1 +5 -2
+2 +5 -2
0
Cu + 2AgNO3  Cu(NO3)2 + 2Ag
6. Add half-reactions; Transfer coefficients to
skeleton equation
Cu  Cu+2 + 2e+______________________
2Ag+1 + 2e-  2Ag
Cu + 2Ag+1 + 2e-  2Ag + Cu+2 + 2eCu + 2Ag+1  2Ag + Cu+2
Cu + 2AgNO3  Cu(NO3)2 + 2 Ag
```
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