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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 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) + 8 H+(aq) + 5 e- Mn2+(aq) + 4 H2O(l) + 1.51
Cl2(g) + 2 e- 2 Cl-(aq) + 1.36
Cr2O72-(aq) + 14 H+(aq) + 6 e- 2 Cr3+(aq) + 7 H2O(l) + 1.33
Br2(g) + 2 e- 2 Br-(aq) + 1.09
Ag+(aq) + e- Ag(s) + 0.80
Fe3+(aq) + e- Fe2+(aq) + 0.77
MnO4-(aq) + e- MnO4
2-(aq) + 0.56
I2(g) + 2 e- 2 I-(aq) + 0.54
Cu2+(aq) + 2 e- Cu(s) + 0.34
Hg2Cl2(aq) + 2 e- 2 Hg(l) + 2 CI-(aq) + 0.27
AgCl(s) + e- Ag(s) + Cl-(aq) + 0.22
2 H+(aq) + 2 e- H2(g) 0.00
Pb2+(aq) + 2 e- Pb(s) - 0.13
Sn2+(aq) + 2 e- Sn(s) - 0.14
V3+(aq) + e- V2+(aq) - 0.26
Ni2+(aq) + 2 e- Ni(s) - 0.25
Fe2+(aq) + 2 e- Fe(s) - 0.44
Zn2+(aq) + 2 e- Zn(s) - 0.76
Al3+(aq) + 3 e- Al(s) - 1.66
Mg2+(aq) + 2 e- Mg(s) - 2.36
Na+(aq) + e- Na(s) - 2.71
Ca2+(aq) + 2 e- Ca(s) - 2.87
K+(aq) + e- K(s) - 2.93
Increasingreducing
power
Increasingoxidising
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
– 0.76 V
–ve electrode
Zn2+ + 2 e- Zn
+ 0.34 V
+ve electrode
Cu2+ + 2 e- Cu
+ 1.10 V
e–
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?
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 workit gets the electrons by reversing the half reaction with the lower E° value
therefore Cu2+(aq) ——> Cu(s) and
Sn(s) ——> Sn2+(aq)
the overall reaction is Cu2+(aq) + Sn(s) ——> Sn2+(aq) + Cu(s)
the cell voltage is the difference in E° values... (+0.34) - (-0.14) = + 0.48V
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
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 Sn(s) ——> Sn2+(aq)
The cell voltage will be the differencein E° values and will be positive... (+0.34) - (- 0.14) = + 0.48V
If this is the equation you want then it will be spontaneousIf 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) + 2e¯ ——> Cu(s)
Sn(s) ——> Sn2+(aq) + 2e¯
Find the electrode potentials Cu2+(aq) + 2e¯ Cu(s) ; E° = +0.34V
and the usual equations Sn2+(aq) + 2e¯ Sn(s) ; E° = - 0.14V
Reverse one equation and its sign Sn(s) ——> Sn2+(aq) + 2e¯ ; E° = +0.14V
Combine the two half equations Sn(s) + Cu2+(aq) ——> 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
• Predicting redox reactions
• 5.3 exercise 2
– + 0
– 0.76 V
–ve electrode
Zn2+ + 2 e- Zn
– 0.25 V
+ve electrode
Ni2+ + 2 e- Ni
+ 0.51 V
e–
Ni2+ + Zn → Ni + Zn2+
PREDICTING REDOX REACTIONS – Q1
+ 0
+ 0.34 V
–ve electrode
Cu2+ + 2 e- Cu
+ 0.80 V
+ve electrode
Ag+ + e- Ag
+ 0.46 V
e–
2 Ag+ + Cu → 2 Ag + Cu2+
PREDICTING REDOX REACTIONS – Q2
0
– 2.36 V
–ve electrode
Mg2+ + 2 e- Mg
– 0.26 V
+ve electrode
V3+ + e- V2+
+ 2.10 V
e–
Mg(s)|Mg2+(aq)||V3+(aq),V2+(aq)|Pt(s)
PREDICTING REDOX REACTIONS – Q3 a
–
YES: Mg reduces V3+ to V2+
0
+ 0.77 V
–ve electrode
Fe3+ + e- Fe2+
+ 1.36 V
+ve electrode
Cl2 + 2 e- 2 Cl-
+ 0.59 V
e–
PREDICTING REDOX REACTIONS – Q3 b
+
NO: Cl- won’t reduce Fe3+ to Fe2+
0
+ 1.09 V
–ve electrode
+ 1.36 V
+ve electrode
Cl2 + 2 e- 2 Cl-
+ 0.27 V
e–
PREDICTING REDOX REACTIONS – Q3 c
+
YES: Cl2 oxidises Br- to Br2 Br2 + 2 e- 2 Br-
Pt(s)|Br-(aq),Br2(aq)||Cl2(g)|Cl-(aq)|Pt(s)
0
– 0.14 V
–ve electrode
Sn2+ + 2 e- Sn
+ 0.77 V
+ve electrode
Fe3+ + e- Fe2+
+ 0.91 V
e–
PREDICTING REDOX REACTIONS – Q3 d
–
YES: Sn reduces Fe3+ to Fe2+
+
Sn(s)|Sn2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s)
0
+ 1.33 V
–ve electrode
Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O
+ 1.36 V
+ve electrode
Cl2 + 2 e- 2 Cl-
+ 0.03 V
e–
PREDICTING REDOX REACTIONS – Q3 e
+
NO: H+/Cr2O72- won’t oxidise Cl- to
Cl2
0
+ 1.36 V
–ve electrode
MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O
+ 1.51 V
+ve electrode
Cl2 + 2 e- 2 Cl-
+ 0.03 V
e–
PREDICTING REDOX REACTIONS – Q3 f
+
YES: H+/MnO4- oxidises Cl- to Cl2
Pt(s)|Cl-(aq)|Cl2(g)||MnO4- (aq),H+(aq),Mn2+(aq)|
Pt(s)
0
– 0.44 V
–ve electrode
Fe2+ + 2 e- Fe
0.00 V
+ve electrode
2 H+ + 2 e- H2
+ 0.44 V
e–
PREDICTING REDOX REACTIONS – Q3 g
–
YES: H+ oxidises Fe to Fe2+
Fe(s)|Fe2+(aq)||H+(aq)|H2(g)|Pt(s)
0
0.00 V
–ve electrode
Cu2+ + 2 e- Cu
+ 0.34 V
+ve electrode
2 H+ + 2 e- H2
+ 0.34 V
e–
PREDICTING REDOX REACTIONS – Q3 h
+
NO: H+ won’t oxidise Cu to Cu2+
0
+ 1.36 V
MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O
+ 1.51 V
Cl2 + 2 e- 2 Cl-
PREDICTING REDOX REACTIONS – Q4
+
+ 1.33 V Cr2O7
2- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O
+ 0.77 V Fe3+ + e- Fe2+
YES
NO
NO
+ 0
? V
–ve electrode
Be2+ + 2 e- Be
+ 0.34 V
+ve electrode
Cu2+ + 2 e- Cu
+ 2.19 V
e–
Be2+ + Cu → Be + Cu2+
PREDICTING REDOX REACTIONS – Q5a
2.19 = 0.34 - Eleft
Eleft = 0.34 – 2.19 = – 1.85 V
– 0
? V
–ve electrode
Th4+ + 4 e- Th
+ 0.00 V
+ve electrode
1.90 V
e–
4 H+ + Th → 2 H2 + Th4+
PREDICTING REDOX REACTIONS – Q5b
When using SHE
E = cell emf = – 1.90 V
2 H+ + 2 e- H2
0
0.00 V
–ve electrode
+ 1.09 V
+ve electrode
+ 1.09 V
e–
PREDICTING REDOX REACTIONS – Q6a
+
Br2 + 2 e- 2 Br-
Pt(s)|H2(g)|H+(aq)||Br2(aq),Br-(aq)|Pt(s)
2 H+ + 2 e- H2 H2 + Br2 → 2 H+ + 2 Br-
0
+ 0.34 V
–ve electrode
+ 0.77 V
+ve electrode
+ 0.43 V
e–
PREDICTING REDOX REACTIONS – Q6b
+
Fe3+ + e- Fe2+
Cu(s)|Cu2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s)
Cu2+ + 2 e- Cu 2 Fe3+ + Cu → 2 Fe2+ + Cu2+
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