chem1612 - pharmacy week 9: galvanic cells dr. siegbert schmid school of chemistry, rm 223 phone:...
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CHEM1612 - PharmacyWeek 9: Galvanic Cells
Dr. Siegbert Schmid
School of Chemistry, Rm 223
Phone: 9351 4196
E-mail: [email protected]
Unless otherwise stated, all images in this file have been reproduced from:
Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, John Wiley & Sons Australia, Ltd. 2008
ISBN: 9 78047081 0866
Lecture 25-3
Electrochemistry Blackman, Bottle, Schmid, Mocerino & Wille:
Chapter 12, Sections 4.8 and 4.9
Key chemical concepts: Redox and half reactions Cell potential Voltaic and electrolytic cells Concentration cells
Key Calculations: Calculating cell potential Calculating amount of product for given current Using the Nernst equation for concentration cells
Lecture 25-4
The measured voltage across the cell is called cell potential (Ecell). This driving force for the reaction is also called ELECTROMOTIVE
FORCE (emf) of the cell.
Every galvanic cell has a different cell potential. How can we measure the cell potential relative to each species? How can we tabulate cell potentials?
Electromotive Force
Zn(s) + Cu2+(aq) Zn2+
(aq) + Cu(s)
Lecture 25-5
The electromotive force cannot be measured absolutely for one element (we cannot measure EZn and ECu in isolation).
Only differences in potential have meaning (ΔE).
We assign a “half-cell” potential to each “half-reaction”, and then add up two half-reactions to get the full reaction and full potential.
We choose as a standard the reaction of hydrogen :
2H+ (aq) + 2e– H2 (g) E0 = 0.00 V
(1 atm H2, [H+] = 1 M, at all temperatures)
Standard Reduction Potential
Indicates standardconditions
Lecture 25-6
Determining E0cell
We measure half-cell potentials relative to this reference hydrogen (H+/H2) electrode, which has E0
ref = 0.00 V. We build voltaic cells that have this reference half-cell and another
half-cell whose potential is unknown, E0unknown.
We define: E0cell = E0
cathode – E0anode
When H2 is oxidised, the reference electrode is the anode, so:
E0cell = E0
cathode – E0ref = E0
unknown – 0.00 = E0unknown
When H+ is reduced, the reference electrode is the cathode, so:
E0cell = E0
ref – E0anode = 0.00 - E0
unknown = -E0unknown
Lecture 25-7
Standard
Hydrogen
Electrode
Experiment 1: Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
0.76 V
Potential(V)
all reagents at standard concentration of 1.0 M
Demo: Potential of cell Zn/H2
Zn(s)|Zn2+||H+(aq)|H2(g)|Pt
We measure E0cell = 0.76 V;
Since E0cell = 0 - E0
unknown
Then
E0Zn = -0.76 V
Lecture 25-8
Cu2+(s) + H2(g) 2H+(aq) + Cu(s) Experiment 2:
+0.34 VPotential(V)
Demo: Potential of cell Cu/H2
H2(g)| H+(aq)|| Cu2+(s)|Cu(s)| Pt
all reagents at standard concentration of 1.0 M
We measure E0cell =0.34 V
Since E0cell = E0
unknown
Then
E0Cu = 0.34 V
Lecture 25-9
E0Cu
0.34 V
E0Zn -0.76 V
1.10 V
Potential(V)Experiment 3: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Demo: Potential of Zn/Cu cell
We measure a potential
E0cell = 1.1 V.
Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
E0cell = E0
cathode – E0anode = 0.34 – (–0.76)= 0.34 + 0.76 = 1.10 V
Lecture 25-10
Cell Potential
all reagents at standard concentration of 1.0 M
- Experiment 1: Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
Experiment 3: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Cu2+(aq) + H2(g) Cu(s) + 2H+(aq)Experiment 2:
-(-0.76 V)
0.34 V
= 1.10 V
Potential(V)
Conclusion 2:
You can add cell potentials as you add chemical reactions.
Conclusion 2:
You can add cell potentials as you add chemical reactions.
Conclusion 1:
Reverse the reaction – reverse the sign of the potential.
Lecture 25-11
We measured for the reaction
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
A cell potential Ecell = ΔE = 1.1 V
Measured in volts, V = J·C–1
Moving one Coulomb of charge from Zn to Cu2+ releases 1.1 J of energy
Electromotive Force ΔE
Lecture 25-12
We define the standard reduction potential E0 as the potential of each redox couple when all reagents are in the standard state, i.e. 1 M concentration, 298 K, and gases at 1 atm pressure.
The standard reduction potentials are tabulated.
In data tables all half reactions are written as reductions.
Fe3+ + e– Fe2+ E0 = 0.77 V
Cu2+ + 2e– Cu E0 = 0.33 V
2H+ + 2e– H2 (g) E0 = 0.00 V
Zn2+ + 2e– Zn E0 = –0.76 V
The higher E (more positive), the greater the tendency to acquire electrons (be reduced).
Standard Reduction Potentials
Lecture 25-13
Reduction potential tableHalf-reaction Half-cell potential (V)
Au3+(aq) + 3e- Au(s) +1.50
Cl2(g) + 2e- 2Cl- (aq) +1.36
O2(g) + 4H+(aq) + 4e- 2H2O(l) +1.23
Ag+(aq) + e- Ag(s) +0.80
Cu2+(aq) + 2e- Cu(s) +0.34
2H+(aq) + 2e- H2(g) 0.00*
Sn2+(aq) + 2e- Sn(s) -0.14
Fe2+(aq) + 2e- Fe(s) -0.44
Zn2+(aq) + 2e- Zn(s) -0.76
2H2O(l) + 2e- H2(g) + 2OH-(aq) -0.83
Mg2+(aq) + 2e- Mg(s) -2.37
* by definition
Strong oxidising
agent
Weak oxidising
agent
Strong reducing
agent
Weak reducing
agent
No/slow oxidation by H+ due to anover-potential
Lecture 25-14
Using the redox potential tables
1. Write down the two half-reactions.
2. Work out which is the oxidation and which is the reduction half-reaction.
3. Balance the electrons.
4. Add up the half-reactions to get full reaction.
5. Add up half-cell potentials to get E 0.
6. A spontaneous (working) voltaic cell, ALWAYS has a positive cell potential, E 0.
Lecture 25-15
ApproachCalculate the standard cell potential for a galvanic cell formed by Mg2+(aq)
| Mg(s) in one half-cell and Sn2+(aq) | Sn(s) in the other.
Which reaction is the oxidation and which is the reduction?
Which half reaction is turned around?
Sn2+ + 2e- Sn E0 = -0.14V
Mg2+ + 2e- Mg E0 = -2.37V
In general, you reverse the reaction that appears lower
in the table of standard reduction potentials.
In general, you reverse the reaction that appears lower
in the table of standard reduction potentials.
Lecture 25-16
Example 1Calculate the standard cell potential for a galvanic cell formed by Mg2+(aq)
| Mg(s) in one half-cell and Sn2+(aq) | Sn(s) in the other.
We saw that Mg is a stronger reducing agent than Sn, i.e. it likes to be oxidised. Turn the Mg reaction around to get an oxidation reaction.
Sn2+ + 2e- Sn E0 = -0.14V
Mg2+ + 2e- Mg E0 = -2.37V
Sn2+ + 2e- Sn E 0 = -0.14V
Mg Mg2+ + 2e- E 0 = +2.37V
Mg(s) + Sn2+(aq) Mg2+(aq) + Sn(s) E 0 = +2. 23V
Lecture 25-17
Example 2Calculate the standard cell potential for a galvanic cell with Ag+(aq) | Ag(s) in one half-cell and Cr3+(aq) | Cr(s) in the other.
E0 (Ag+(aq)|Ag(s)) = +0.80 V; E0(Cr3+(aq)|Cr(s)) = -0.74V
Cr half reaction is lower (more negative), turn it around…
Cr(s) + 3Ag+(aq) Cr3+(aq) + 3Ag(s) E 0 = +1.54V
Balance the electrons…Note!
Note: E0 does not depend on stoichiometry!
Ag+ + e- Ag E 0 = +0.80V
Cr Cr3+ + 3e- E 0 = +0.74V
3Ag+ + 3e- 3Ag E 0 = +0.80V
Cr Cr3+ + 3e- E 0 = +0.74V
Lecture 25-18
Al + Fe2+ Al3+ + Fe
ONs: Al Al3+
Fe2+ Fe
Oxidation half cell: Al Al3+ + 3e-
Reduction half cell: 2e- + Fe2+ Fe
Combine: 2 x (Al Al3+ + 3e-) 3 x (2e- + Fe2+ Fe)
2Al + 3Fe2+ 2Al3+ + 3Fe
Balancing Redox Equations – Example 1
(oxidation)(reduction)
balance charge with e-
balance charge with e-
Lecture 25-19
Cr2O72- + I- Cr3+ + I2 (in acidic solution)
O.N.s: Cr2O72- (+6) Cr3+ (+3)
I- (-1) I2 (0)
Oxidation half cell: 2I- I2 + 2e-
Reduction half cell: 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
Combine: 3 x (2I- I2 + 2e-)
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
6 I- + 14H+ + Cr2O72- 3I2 + 2Cr3+ + 7H2O
(oxidation)
(reduction)
Balancing Redox Equations – Example 2
Lecture 25-20
HgO + Zn Hg + Zn(OH)2 (in basic solution)
ONs: HgO (+2) Hg (0) Zn (0) Zn2+ (+2)
Oxidation half cell: Zn Zn2+ + 2e-
Reduction half cell: 2e- + H2O + HgO Hg + 2OH-
(Add OH- and water to neutralise the charge and balance O and H)
Combine: Zn Zn2+ + 2e-
2e- + H2O + HgO Hg + 2OH-
Zn + H2O + HgO Zn2+ + Hg + 2OH-
Balancing Redox Equations – Example 3
(oxidation)
(reduction)
Lecture 25-21
Step 1 Assign O.N. to all elements.
Step 2 Identify the species being reduced/oxidised.
Step 3 Calculate and balance the gain/loss electrons.
Step 4 Balance the number of all the atoms other than O and H.
Step 5 Balance O by adding H2O to either side as required.
Balance H and charges by adding H+ to either side as required.
Step 6 If basic conditions are specified, add OH– as required.
(H+ and OH- cannot be present at the same time, they will convert into H2O).
Finally, check that the whole reaction is balanced, i.e
that the charges and the moles of reactants and products are balanced.
Balancing Redox Equations – oxidation number method
Lecture 25-22
Example 1
Balance the equation for the oxidation of HCl by MnO4 –
MnO4– + HCl Mn2+
+ Cl2
Answer:
2 MnO4–(aq) + 10 HCl(aq) + 6 H+(aq) 2 Mn2+(aq) + 5 Cl2(aq) + 8 H2O(l)
Lecture 25-23
Example 2
Balance the reaction of oxidation of H2SO3 by Cr2O72-:
Cr2O72- + H2SO3 Cr3+ + SO4
2-
Answer: Cr2O72- + 3 H2SO3 + 2 H+ → 2 Cr3+ + 3 SO4
2- + 4H2O
Lecture 25-24
Example 3
Balance the reaction between NaMnO4 and Na2C2O4:
MnO4- + C2O4
2- MnO2 + CO32-
Answer: 2 MnO4- + 3 C2O4
2-+ 2 H2O → 2 MnO2 +6 CO32- + 4 H+
Lecture 25-25
Summary
CONCEPTS Galvanic/Voltaic cells Cell notation Standard Reduction Potentials Balancing Redox Equations
CALCULATIONS Work out cell potential from reduction potentials
Lecture 25-26
Half-cell standard reduction potentials