predicting the direction of redox reactions know that standard electrode potentials can be listed...
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![Page 1: Predicting the direction of redox reactions Know that standard electrode potentials can be listed as an electrochemical series. Use E values to predict](https://reader035.vdocuments.site/reader035/viewer/2022062300/56649cc15503460f94988af2/html5/thumbnails/1.jpg)
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.
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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
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GOLDEN RULE
The more +ve electrode gains electrons
(+ charge attracts electrons)
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Electrodes with negative emf are better at releasing electrons (better reducing agents).
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– + 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+
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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
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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
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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
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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
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• Predicting redox reactions
• 5.3 exercise 2
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– + 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
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+ 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
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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+
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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+
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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)
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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)
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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
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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)
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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)
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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+
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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
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+ 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
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– 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
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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-
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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+