cell potential
DESCRIPTION
Cell Potential. L.O.: Construct redox equations using half-equations or oxidation numbers. Describe how to make an electrochemical cell. An oxidation number is a measure of the number of electrons that an atom uses to bond with atoms of another element. - PowerPoint PPT PresentationTRANSCRIPT
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Cell Potential
L.O.: Construct redox equations using half-equations or oxidation numbers. Describe how to make an electrochemical cell.
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An oxidation number is a measure of the number of electrons that an atom uses to bond with atoms of another element.
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Each element in a compound is given an oxidation number.
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Species Oxidation number
examples
Uncombined element
0 C, Na, O2, H2, P4, Cl2
Combined oxygen
-2 H2O, CaO
Combined Hydrogen
+1 NH3, H2S
Simple ion Charge on ion Na+, Mg2+, Cl-1
Combined fluorine
-1 NaF, CaF2
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Species Oxidation number
examples
Combined group 1
+1 NaCl
Combined group 2
+2 MgCl2
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The sum of the oxidation numbers must equal the overall charge.
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Oxidation states.
•F = -1
•Cl = -1
•O = -2
•H = +1
Mg in MgCl2?
? - 2 = 0
? = +2
Put the + in!
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No need to balance Mn; equal numbers
BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
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BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
Overall charge on MnO4¯ is -1; sum of the OS’s of all atoms must add up to -1
Oxygen is in its usual oxidation state of -2; four oxygen atoms add up to -8To make the overall charge -1, Mn must be in oxidation state +7 ... [+7 + (4x -2) = -1]
1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2
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BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
The oxidation states on either side are different; +7 —> +2 (REDUCTION)To balance; add 5 negative charges to the LHS [+7 + (5 x -1) = +2]You must ADD 5 ELECTRONS to the LHS of the equation
1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+
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BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
Total charges on either side are not equal; LHS = 1- and 5- = 6-RHS = 2+
Balance them by adding 8 positive charges to the LHS [ 6- + (8 x 1+) = 2+ ]You must ADD 8 PROTONS (H+ ions) to the LHS of the equation
1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+
Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+
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Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+
Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+
Step 5 MnO4¯ + 5e¯ + 8H+ ———> Mn2+ + 4H2O now balanced
BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
Everything balances apart from oxygen and hydrogen O LHS = 4 RHS = 0H LHS = 8 RHS = 0
You must ADD 4 WATER MOLECULES to the RHS; the equation is now balanced
1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side
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5.3 Exercise 1
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BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
REMINDER1 Work out the formula of the species before and after the change; balance if required2 Work out the oxidation state of the element before and after the change3 Add electrons to one side of the equation so that the oxidation states balance4 If the charges on all the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges5 If the equation still doesn’t balance, add sufficient water molecules to one side
Q. Balance the following half equations...
Na —> Na+
Fe2+ —> Fe3+
I2 —> I¯
C2O42- —> CO2
H2O2 —> O2
H2O2 —> H2O
NO3- —> NO
NO3- —> NO2
SO42- —> SO2
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BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
Q. Balance the following half equations...
Na —> Na+ + e-
Fe2+ —> Fe3+ + e-
I2 + 2e- —> 2I¯
C2O42- —> 2CO2 + 2e-
H2O2 —> O2 + 2H+ + 2e-
H2O2 + 2H+ + 2e- —> 2H2O
NO3- + 4H+ + 3e- —> NO + 2H2O
NO3- + 2H+ + e- —> NO2
+ H2O
SO42- + 4H+ + 2e- —> SO2 + 2H2O
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COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows...
Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides
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COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows...
Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides
The reaction between manganate(VII) and iron(II)
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COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
The reaction between manganate(VII) and iron(II)
Step 1 Fe2+ ——> Fe3+ + e¯ OxidationMnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction
A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows...
Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides
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COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
The reaction between manganate(VII) and iron(II)
Step 1 Fe2+ ——> Fe3+ + e¯ OxidationMnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction
Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1
A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows...
Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides
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COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
The reaction between manganate(VII) and iron(II)
Step 1 Fe2+ ——> Fe3+ + e¯ OxidationMnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction
Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1
Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯
A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows...
Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides
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COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
The reaction between manganate(VII) and iron(II)
Step 1 Fe2+ ——> Fe3+ + e¯ OxidationMnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction
Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1
Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯
Step 4 MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+
A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows...
Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides
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The reaction between manganate(VII) and iron(II)
Step 1 Fe2+ ——> Fe3+ + e¯ OxidationMnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction
Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1
Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯
Step 4 MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+
COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
SUMMARY
A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows...
Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides
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A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows...
Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides
COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
Q. Construct balanced redox equations for the reactions between...
Mg and H+
Cr2O72- and Fe2+
H2O2 and MnO4¯
C2O42- and MnO4¯
S2O32- and I2
Cr2O72- and I¯
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Mg ——> Mg2+ + 2e¯ (x1)H+ + e¯ ——> ½ H2 (x2)
Mg + 2H+ ——> Mg2+ + H2
Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O (x1)
Fe2+ ——> Fe3+ + e¯ (x6)
Cr2O72- + 14H+ + 6Fe2+ ——> 2Cr3+ + 6Fe2+ + 7H2O
MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O (x2)
H2O2 ——> O2 + 2H+ + 2e¯ (x5)
2MnO4¯ + 5H2O2 + 6H+ ——> 2Mn2+ + 5O2 + 8H2O
MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O (x2)
C2O42- ——> 2CO2 + 2e¯ (x5)
2MnO4¯ + 5C2O42- + 16H+ ——> 2Mn2+ + 10CO2 + 8H2O
2S2O32- ——> S4O6
2- + 2e¯ (x1)
½ I2 + e¯ ——> I¯ (x2)
2S2O32- + I2 ——> S4O6
2- + 2I¯
Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O (x1)
½ I2 + e¯ ——> I¯ (x6)
Cr2O72- + 14H+ + 3I2 ——> 2Cr3+ + 6I ¯ + 7H2O
BALANCING
REDOX
EQUATIONS
ANSWERS
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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.
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Half-cell: an elements in two oxidation states.
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Zn2+(aq) + 2 e– Zn(s)
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Zn2+(aq) + 2e¯ -> Zn(s) E° = - 0.76V
The electrode potential of the half cell indicates its tendency to lose or gain electrons.
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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
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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)
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S tandard H ydrogen E lectrode
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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
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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
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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)
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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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Emf = Eright - Eleft
ELECTRODE POTENTIALS – Q1
- 2.71 = Eright - 0
Eright = - 2.71 V
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Emf = Eright - Eleft
ELECTRODE POTENTIALS – Q2
Emf = - 0.44 - 0.22
Emf = - 0.66 V
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Emf = Eright - Eleft
ELECTRODE POTENTIALS – Q3
Emf = - 0.13 - (-0.76)
Emf = + 0.63 V
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Emf = Eright - Eleft
ELECTRODE POTENTIALS – Q4
+1.02 = +1.36 - Eleft
Eleft = + 1.36 - 1.02 = +0.34 V
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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
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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)
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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)