starter find the balanced redox equations for: 1)h 2 o 2 with mno 4 - to mn 2+ and o 2 2)cr 2 o 7 2-...
TRANSCRIPT
Starter
Find the balanced redox equations for:1) H2O2 with MnO4
- to Mn2+ and O2
2) Cr2O72- with I2 to give I- and Cr3+
Extension:S2O3
2- and I2 to give I- and S4O62-
Starter
Find the balanced redox equations for:1) H2O2 with MnO4
- to Mn2+ and O2
MnO4¯ + 5e¯ + 8H+ —> Mn2+ + 4H2O(x2)
H2O2 ——> O2 + 2H+ + 2e¯ (x5)
2MnO4¯ + 5H2O2 + 6H+ —> 2Mn2+ + 5O2 + 8H2O
Starter
Find the balanced redox equations for:Cr2O7
2- with I2 to give I- and Cr3+
Cr2O72- + 14H+ + 6e¯ —> 2Cr3+ + 7H2O (x1)
½ I2 + e¯ —> I¯ (x6)
Cr2O72- + 14H+ + 3I2 —> 2Cr3+ + 6I ¯ +
7H2O
Starter
S2O32- and I2 to give I- and S4O6
2-
• 2S2O32- ——> S4O6
2- + 2e¯ (x1)
• ½ I2 + e¯ ——> I¯ (x2)
• 2S2O32- + I2 ——> S4O6
2- + 2I¯
Cell Potential
L.O.: Define the term standard electrode potential.Describe how to make an electrochemical cell. Calculate the standard cell potential of a cell.
• Homework
Electrode Potentials
know the IUPAC convention for writing half-equations for electrode reactions.
Know and be able to use the conventional representation of cells.
Know that standard electrode potential, E , refers to conditions of 298 K, 100 kPa and 1.00 mol dm−3 solution of ions.
Half-cell: an element in two oxidation states.
Zn2+(aq) + 2 e– Zn(s)
Zn2+(aq) + 2e¯ -> Zn(s) E° = - 0.76V
The electrode potential of the half cell indicates its tendency to lose or gain electrons.
The standard electrode potential of a half-cell, is the e.m.f of a cell compared with a standard hydrogen half-cell, measured at 298 K with a solution concentration of 1 mol dm-3 and a gas pressure of 100KPa
Standard Conditions
Concentration 1.0 mol dm-3 (ions involved in ½ equation)
Temperature 298 K
Pressure 100 kPa (if gases involved in ½ equation)
Current Zero (use high resistance voltmeter)
S tandard H ydrogen E lectrode
Zn
Zn Zn2+ + 2 e-
oxidation
Cu2+ + 2 e- Cureduction
- electrode
anodeoxidation
+ electrodecathode
reductionelectron flow
At this electrode the metal loses
electrons and so is oxidised to metal
ions.
These electrons make the electrode
negative.
At this electrode the metal ions gain
electrons and so is reduced to metal
atoms.
As electrons are used up, this makes the electrode positive.
Cu
Emf = E = E(positive
terminal) - E(negative terminal)
H2 at 100 kPa
o
o
o
o
o
o
o
o
o
o
o
o
salt bridge
1.0 M H+(aq)
Pt
temperature= 298 K
1.0 M Cu2+(aq)
V
Cu
high resistancevoltmeter
H2 at 100 kPa
o
o
o
o
o
o
o
o
o
o
o
o
salt bridge
1.0 M H+(aq)
Pt
temperature= 298 K
1.0 M Cu2+(aq)
V
Cu
high resistancevoltmeter
Pt(s) | H2(g) | H+(aq) || Cu2+(aq) | Cu(s)
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).
• A2.CHEM5.3.003
5.3 EXERCISE 2 - electrochemical cells
Emf = Eright - Eleft
ELECTRODE POTENTIALS – Q1
- 2.71 = Eright - 0
Eright = - 2.71 V
Emf = Eright - Eleft
ELECTRODE POTENTIALS – Q2
Emf = - 0.44 - 0.22
Emf = - 0.66 V
Emf = Eright - Eleft
ELECTRODE POTENTIALS – Q3
Emf = - 0.13 - (-0.76)
Emf = + 0.63 V
Emf = Eright - Eleft
ELECTRODE POTENTIALS – Q4
+1.02 = +1.36 - Eleft
Eleft = + 1.36 - 1.02 = +0.34 V
Emf = Eright - Eleft
ELECTRODE POTENTIALS – Q5
a) Emf = + 0.15 - (-0.25) = +0.40 Vb) Emf = + 0.80 - 0.54 = +0.26 Vc) Emf = + 1.07 - 1.36 = - 0.29 V
Emf = Eright - Eleft
ELECTRODE POTENTIALS – Q6
a) Eright = +2.00 - 2.38 = - 0.38 V Ti3+(aq) + e- Ti2+(aq)
b) Eleft = -2.38 - 0.54 = - 2.92 V
K+(aq) + e- K(aq)c) Eright = - 3.19 + 0.27 = - 2.92 V Ti3+(aq) + e- Ti2+(aq)
ELECTRODE POTENTIALS – Q7
Emf = -0.76 - (-0.91) = +0.15 V
a) Cr(s) | Cr2+(aq) || Zn2+(aq) | Zn(s)
Emf = +0.77 - 0.34 = +0.43 V
b) Cu(s) |Cu2+(aq)|| Fe3+(aq),Fe2+(aq)| Pt(s)
Emf = +1.51 – 1.36 = +0.15 V
c) Pt(s) | Cl-(aq)| Cl2(g) || MnO4-(aq),H+(aq),Mn2+(aq)| Pt(s)