Predicting the direction of
redox reactions
Know that standard electrode potentials
can be listed as an electrochemical series.
 Use E values to predict the direction of
simple redox reactions and to calculate
the e.m.f. of a cell.
Standard electrode potentials
Increasing
oxidising
power
E/V
F2(g) + 2 e-

2 F-(aq)
+ 2.87
MnO42-(aq) + 4 H+(aq) + 2 e-

MnO2(s) + 2 H2O(l)
+ 1.55
MnO4-(aq)
Cr2O72-(aq)
+
-
+ 8 H (aq) + 5 e
Cl2(g) + 2 e+
-
+ 14 H (aq) + 6 e

2+
Mn (aq) + 4 H2O(l)
+ 1.51

2 Cl-(aq)
+ 1.36

3+
+ 1.33
-
2 Cr (aq) + 7 H2O(l)
-
Br2(g) + 2 e

2 Br (aq)
+ 1.09
Ag+(aq) + e-

Ag(s)
+ 0.80
3+
-

Fe (aq)
+ 0.77
-

MnO42-(aq)
+ 0.56
-
Fe (aq) + e
MnO4-(aq)
2+
+ e

-
2 I (aq)
+ 0.54
Cu (aq) + 2 e

Cu(s)
+ 0.34
Hg2Cl2(aq) + 2 e-

2 Hg(l) + 2 CI-(aq)
I2(g) + 2 e
2+

Ag(s) + Cl (aq)
-
2 H (aq) + 2 e

H2(g)
0.00
Pb2+(aq) + 2 e-
AgCl(s) + e
+
-
+ 0.27
-
+ 0.22

Pb(s)
- 0.13
-
Sn (aq) + 2 e

Sn(s)
- 0.14
V3+(aq) + e-

V2+(aq)
- 0.26
2+
-

Ni(s)
- 0.25
2+
-
Fe (aq) + 2 e

Fe(s)
- 0.44
Zn2+(aq) + 2 e-
2+
Ni (aq) + 2 e

Zn(s)
- 0.76
-
Al (aq) + 3 e

Al(s)
- 1.66
Mg2+(aq) + 2 e-
3+

Mg(s)
- 2.36
-
Na (aq) + e

Na(s)
- 2.71
2+
-
Ca (aq) + 2 e

Ca(s)
- 2.87
K+(aq) + e-

K(s)
- 2.93
+
Increasing
reducing
power
GOLDEN RULE
The more +ve electrode
gains electrons
(+ charge attracts electrons)
Electrodes with negative emf are better at
releasing electrons (better reducing
agents).
–
0
–ve electrode
+
+ve electrode
e–
+ 1.10 V
+ 0.34 V
Cu2+ + 2 e-  Cu
– 0.76 V
Zn2+ + 2 e-  Zn
Cu2+ + Zn → Cu + Zn2+
USE OF Eo VALUES - WILL IT WORK?
E° values
Can be used to predict the feasibility of redox and cell reactions
In theory ANY REDOX REACTION WITH A POSITIVE E° VALUE WILL WORK
An equation with a more positive E° value reverse a less positive one
USE OF Eo VALUES - WILL IT WORK?
An equation with a more positive E° value reverse a less positive one
What happens if an Sn(s) / Sn2+(aq) and a Cu(s) / Cu2+(aq) cell are connected?
Write out the equations
Cu2+(aq) + 2e¯
Cu(s)
; E° = +0.34V
Sn2+(aq) + 2e¯
Sn(s)
; E° = -0.14V
the half reaction with the more positive E° value is more likely to work
it gets the electrons by reversing the half reaction with the lower E° value
Cu2+(aq) ——> Cu(s)
therefore
Sn(s)
the overall reaction is
Cu2+(aq)
+ Sn(s)
——>
and
Sn2+(aq)
——> Sn2+(aq) + Cu(s)
the cell voltage is the difference in E° values... (+0.34) - (-0.14) = + 0.48V
USE OF Eo VALUES - WILL IT WORK?
An equation with a more positive E° value reverse a less positive one
Will this reaction be spontaneous?
Sn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s)
Write out the appropriate equations
Cu2+(aq) + 2e¯
Cu(s) ; E° = +0.34V
as reductions with their E° values
Sn2+(aq) + 2e¯
Sn(s) ; E° = - 0.14V
The reaction which takes place will involve the more positive one reversing the other
i.e.
Cu2+(aq) ——> Cu(s)
and
The cell voltage will be the difference
in E° values and will be positive...
Sn(s)
——>
Sn2+(aq)
(+0.34) - (- 0.14) = + 0.48V
If this is the equation you want then it will be spontaneous
If it is the opposite equation (going the other way) it will not be spontaneous
USE OF Eo VALUES - WILL IT WORK?
An equation with a more positive E° value reverse a less positive one
Will this reaction be spontaneous?
Sn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s)
Split equation into two half equations
Cu2+(aq)
Sn(s)
Find the electrode potentials Cu2+(aq) + 2e¯
+ 2e¯ ——>
——> Sn2+(aq)
Cu(s) ;
Cu(s)
+ 2e¯
E° = +0.34V
and the usual equations
Sn2+(aq) + 2e¯
Sn(s) ;
Reverse one equation and its sign
Sn(s) ——> Sn2+(aq)
Combine the two half equations
Sn(s) + Cu2+(aq)
+ 2e¯ ; E° = +0.14V
——>
Sn2+(aq) + Cu(s)
Add the two numerical values (+0.34V) + (+ 0.14V) = +0.48V
If the value is positive the reaction will be spontaneous
E° = - 0.14V
• Predicting redox reactions
• 5.3 exercise 2
PREDICTING REDOX REACTIONS – Q1
–
0
–ve electrode
+ve electrode
e–
+ 0.51 V
– 0.25 V
Ni2+ + 2 e-  Ni
– 0.76 V
Zn2+ + 2 e-  Zn
Ni2+ + Zn → Ni + Zn2+
+
PREDICTING REDOX REACTIONS – Q2
0
+
–ve electrode
+ve electrode
e–
+ 0.46 V
+ 0.80 V
Ag+ + e-  Ag
+ 0.34 V
Cu2+ + 2 e-  Cu
2 Ag+ + Cu → 2 Ag + Cu2+
PREDICTING REDOX REACTIONS – Q3 a
Mg(s)|Mg2+(aq)||V3+(aq),V2+(aq)|Pt(s)
–
0
–ve electrode
+ve electrode
e–
+ 2.10 V
– 0.26 V
V3+ + e-  V2+
– 2.36 V
Mg2+ + 2 e-  Mg
YES: Mg reduces V3+ to V2+
PREDICTING REDOX REACTIONS – Q3 b
+
0
–ve electrode
+ve electrode
e–
+ 0.59 V
+ 1.36 V
Cl2 + 2 e-  2 Cl-
+ 0.77 V
Fe3+ + e-  Fe2+
NO: Cl- won’t reduce Fe3+ to Fe2+
PREDICTING REDOX REACTIONS – Q3 c
0
Pt(s)|Br-(aq),Br2(aq)||Cl2(g)|Cl-(aq)|Pt(s)
–ve electrode
+
+ve electrode
e–
+ 0.27 V
+ 1.36 V
Cl2 + 2 e-  2 Cl-
+ 1.09 V
Br2 + 2 e-  2 Br-
YES: Cl2 oxidises Br- to Br2
PREDICTING REDOX REACTIONS – Q3 d
Sn(s)|Sn2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s)
–
+
0
–ve electrode
+ve electrode
e–
+ 0.91 V
+ 0.77 V
Fe3+ + e-  Fe2+
– 0.14 V
Sn2+ + 2 e-  Sn
YES: Sn reduces Fe3+ to Fe2+
PREDICTING REDOX REACTIONS – Q3 e
+
0
–ve electrode
+ve electrode
e–
+ 0.03 V
+ 1.36 V
Cl2 + 2 e-  2 Cl-
+ 1.33 V
Cr2O72- + 14 H+ + 6 e-  2 Cr3+ + 7 H2O
NO: H+/Cr2O72- won’t oxidise Cl- to Cl2
PREDICTING REDOX REACTIONS – Q3 f
Pt(s)|Cl-(aq)|Cl2(g)||MnO4- (aq),H+(aq),Mn2+(aq)|Pt(s)
+
0
–ve electrode
+ve electrode
e–
+ 0.03 V
+ 1.51 V
+ 1.36 V
MnO4- + 8 H+ + 5 e-  Mn2+ + 4 H2O
Cl2 + 2 e-  2 Cl-
YES: H+/MnO4- oxidises Cl- to Cl2
PREDICTING REDOX REACTIONS – Q3 g
Fe(s)|Fe2+(aq)||H+(aq)|H2(g)|Pt(s)
–
0
–ve electrode
+ve electrode
e–
+ 0.44 V
0.00 V
2 H+ + 2 e-  H2
– 0.44 V
Fe2+ + 2 e-  Fe
YES: H+ oxidises Fe to Fe2+
PREDICTING REDOX REACTIONS – Q3 h
0
+
–ve electrode
+ve electrode
e–
+ 0.34 V
+ 0.34 V
Cu2+ + 2 e-  Cu
0.00 V
2 H+ + 2 e-  H2
NO: H+ won’t oxidise Cu to Cu2+
PREDICTING REDOX REACTIONS – Q4
+
0
+ 1.36 V
Cl2 + 2 e-  2 Cl-
+ 0.77 V
+ 1.51 V
Fe3+ + e-  Fe2+
NO
MnO4- + 8 H+ + 5 e-  Mn2+ + 4 H2O
+ 1.33 V
Cr2O72- + 14 H+ + 6 e-  2 Cr3+ + 7 H2O
NO
YES
PREDICTING REDOX REACTIONS – Q5a
0
+
–ve electrode
2.19 = 0.34 - Eleft
+ve electrode
Eleft = 0.34 – 2.19 = – 1.85 V
e–
+ 2.19 V
+ 0.34 V
Cu2+ + 2 e-  Cu
?V
Be2+ + 2 e-  Be
Be2+ + Cu → Be + Cu2+
PREDICTING REDOX REACTIONS – Q5b
–
0
–ve electrode
When using SHE
+ve electrode
E = cell emf = – 1.90 V
e–
1.90 V
+ 0.00 V
2 H+ + 2 e-  H2
?V
Th4+ + 4 e-  Th
4 H+ + Th → 2 H2 + Th4+
PREDICTING REDOX REACTIONS – Q6a
Pt(s)|H2(g)|H+(aq)||Br2(aq),Br-(aq)|Pt(s)
+
0
–ve electrode
+ve electrode
e–
+ 1.09 V
+ 1.09 V
Br2 + 2 e-  2 Br-
0.00 V
2 H+ + 2 e-  H2
H2 + Br2 → 2 H+ + 2 Br-
PREDICTING REDOX REACTIONS – Q6b
Cu(s)|Cu2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s)
0
–ve electrode
+ve electrode
e–
+ 0.43 V
+ 0.77 V
Fe3+ + e-  Fe2+
+ 0.34 V
Cu2+ + 2 e-  Cu
2 Fe3+ + Cu → 2 Fe2+ + Cu2+
+