OFB Chapter 13 13/23/2004
Chapter 13Electrochemistry and Cell Voltage
• 13-1 The Gibbs Function and Cell Voltage
• 13-2 Half Cell Potentials• 13-3 Oxidizing and Reducing
Agents• 13-4 Concentrations and the Nernst
Equations• 13-5 Equilibrium Constants from
Electrochemistry• 13-6 Batteries and Fuel Cells• 13-7 Corrosion and Its Prevention
OFB Chapter 13 23/23/2004
• Important Chapter - ties together several concepts from previous Chapters– Oxidation-Reduction– Electrochemical Cells– Acid-Base– Solubility– Equilibrium constants
• Practical Importance– Energy storage – batteries– Energy conversion – solar cells– Chemical conversion
• E.g., printed circuits – metal deposition• Corrosion• Corrosion prevention
Chapter 13Electrochemistry and Cell Voltage
OFB Chapter 13 33/23/2004
The Gibbs Function and Cell Voltage
Anode
(-)Oxidation
Cathode
(+)Reduction
e- flow →
Q = charge
There exists “potential” or potential difference , called ∆E, between the cells
OFB Chapter 13 43/23/2004
For a Galvanic Cell, a spontaneous reaction takes place.
For the Galvanic cell the cell performs electrical work on the surroundings (acts as a battery)
For an Electrolytic Cell, a non-spontaneous reaction takes place.
For the electrolytic cell, the cell performs electrical work on the system
OFB Chapter 13 53/23/2004
Welec = - Q ∆E
Type Cell ∆E Welec Performs work:
Galvanic on surroundings
Electrolytic on system
OFB Chapter 13 63/23/2004
Welec = - Q ∆E
Welec = - Q ∆E
recall Q = I x t
Substituting gives
W units is Joules ∆E units is Joules/coulomb
OFB Chapter 13 73/23/2004
Consider a Galvanic Cell (recall that this is a spontaneous reaction)
Let’s related ∆G (The Gibbs Function or Gibbs free energy) to Welec
∆G = Welec,max (at constant T,P)
Max. means theoretical max, some energy is lost as heat
Remember Faradays constant?
F = 96, 485 coulombs/moleIf n moles of electrons pass through our Galvanic Cell
∆G = = (at constant T,P)
OFB Chapter 13 83/23/2004
Standard States and Cell Voltages
Standard States usually at 25°C, all reactants and products at standard states.
Example 13-2 and Exercise 13-2 give a straightforward example of calculating ∆G°for a chemical reaction in a cell
TRY THEM
OFB Chapter 13 103/23/2004
13-2 Half Cell Potentials
This section introduces Standard half cell reduction potentials
Tables that list half cell reactions as reductions. (This table from 3rd Ed OFB is slightly different than 4th edition)
OFB Chapter 13 113/23/2004
Standard Reduction potentials E °
Ni2+ + 2e- → Ni (s) -0.257 Volts
Cu2+ + 2e- → Cu (s) +0.345 Volts
Now if we combine these as two separate cells (as we did in chapter 12)
Cu2+ reduction potential is more positive (+0.34) therefore it is reduced (cathode) and Ni is oxidized (anode)
Notice that we want the opposite of the Nickel reduction potential, therefore the minus sign
OFB Chapter 13 123/23/2004
By international convention
2H+ + 2e- → H2 (g)E = 0.0 volts
All reduction potentials can be determined relative to H+
reduction.
OFB Chapter 13 133/23/2004
Chapter 13Electrochemistry and Cell Voltage
• Typical problem (Chapter 13 #10 end of chapter)A Galvanic cell is constructed in which at Pt|Fe2+, Fe3+ half cell is connected to a Cd2+|Cd half cell.(a) By referring to Appendix E, write balanced chemical equations for the half reaction at the anode and the cathode and for the overall reaction(b) Calculate the cell voltage, assuming that all reactions and products are in standard state at 298.15°K.
OFB Chapter 13 143/23/2004
A Galvanic cell is constructed in which at Pt|Fe2+, Fe3+ half cell is connected to a Cd2+|Cd half cell.
(a) By referring to Appendix E, write balanced chemical equations for the half reaction at the anode and the cathode and for the overall reaction
Fe3+ + e- → Fe2+
Cd+2 + 2e- → Cd
Look up Reduction Potentials in Appendix E
For Fe3+ to Fe2+ and Cd+2 to Cd
Need to balance electrons and stoichiometry
OFB Chapter 13 153/23/2004
A Galvanic cell is constructed in which at Pt|Fe2+, Fe3+ half cell is connected to a Cd2+|Cd half cell.
(b) Calculate the cell voltage, assuming that all reactions and products are in standard state at 298.15°K.
Fe3+ reduction potential is more positive (+0.771) therefore it is reduced (cathode) and Cd (-0.403) is oxidized (anode)
Notice that the numbers of electrons appearing in half-equations do not figure in the computation of the potential difference of a cell
OFB Chapter 13 163/23/2004
Oxidizing and Reducing Agents
Supplies ElectronsIncrease
4. Reducing Agent, does the reducing
Picks Up electronsDecrease
3. Oxidizing Agent, does the oxidizing
Gain of ElectronsDecrease2. Reduction
Loss of ElectronsIncrease1. Oxidation
Electron Change
Oxidation Number ChangeTerm
OFB Chapter 13 183/23/2004
Example 13-5
(a) Determine whether 1.00 M H3PO3 (aq) and 1.00 M SO2 (aq) are stable with respect to disproportionation in acidic solution (pH 0 at 25C).
(b) Determine which is the stronger reducing agent under these conditions. Refer to 13-5 for necessary data
OFB Chapter 13 193/23/2004
Example 13-5
(a) Determine whether 1.00 M H3PO3 (aq) and 1.00 M SO2 (aq) are stable with respect to disproportionation in acidic solution (pH 0 at 25C).
(b) Determine which is the stronger reducing agent under these conditions. Refer to 13-5 for necessary data
This is a disproportionation problem, write to get both H3PO3 on the left and then add and calculated ∆E
H3PO3 + 2H+ + 2e-→ H3PO2 + H2O
H3PO3 + H2O → H3PO4 + 2H+ + 2e-
Now Calc. ∆E
OFB Chapter 13 203/23/2004
Example 13-5
(a) Determine whether 1.00 M H3PO3 (aq) and 1.00 M SO2 (aq) are stable with respect to disproportionation in acidic solution (pH 0 at 25C).
Only two half reactions involving H3PO3
H3PO3 + 2H+ + 2e-→ H3PO2 + H2O E= (-0.499) V
H3PO4 + 2H+ + 2e- → H3PO3 + H2O E= (-0.276) V
This is disproportionation, write to get both H3PO3 on the left and then add and calculated ∆E
H3PO3 + 2H+ + 2e-→ H3PO2 + H2O
H3PO3 + H2O → H3PO4 + 2H+ + 2e-
Reduction
Oxidation
OFB Chapter 13 213/23/2004
Why stable?
Recall
For an Electrolytic Cell, a non-spontaneous reaction takes place.
For the electrolytic cell, the cell performs electrical work on the system
Disproportionation is
2 H3PO3 → H3PO2 + H3PO4
OFB Chapter 13 223/23/2004
Example 13-5
(a) Determine whether 1.00 M H3PO3 (aq) and 1.00 M SO2 (aq) are stable with respect to disproportionation in acidic solution (pH 0 at 25C).
Only two half reactions involving SO2
2SO2 + 2H+ + 4e-→ S2O3 -2 + H2O
HSO4- + 3H+ + 2e- → SO2 + 2H2O
Disproportionation of SO2
Rewrite to get both SO2 on the left, balance electrons and stoichiometry and add
OFB Chapter 13 233/23/2004
Example 13-5
(a) Determine whether 1.00 M H3PO3 (aq) and 1.00 M SO2 (aq) are stable with respect to disproportionation in acidic solution (pH 0 at 25C).
∆E = E (cathode) – E (anode)
= 0.40 – (+ 0.172) = +0.228 ∴unstable
Why unstable? It’s a spontaneous reaction.
RecallFor a Galvanic Cell, a spontaneous reaction takes place.
For the Galvanic cell the cell performs electrical work on the surroundings (acts as a battery)
Disproportionation of SO2
4 SO2 + 3 H2O → 2 HSO4- + S2O3
-2 + 4H+
OFB Chapter 13 243/23/2004
Example 13-5
(a) Determine whether 1.00 M H3PO3 (aq) and 1.00 M SO2 (aq) are stable with respect to disproportionation in acidic solution (pH 0 at 25C).
H3PO3 is stable
(non-spontaneous disproportionation)
and
SO2 is unstable
(spontaneous disproportionation)
(b) Determine which is the stronger reducing agent under these conditions. Refer to 13-5 for necessary data
OFB Chapter 13 253/23/2004
Example 13-5
(a) Determine whether 1.00 M H3PO3 (aq) and 1.00 M SO2 (aq) are stable with respect to disproportionation in acidic solution (pH 0 at 25C).
(b) Determine which is the stronger reducing agent under these conditions. Refer to 13-5 for necessary data
Only two half reactions involving SO2
2SO2 + 2H+ + 4e-→ S2O3 -2 + H2O E= +0.40 V
HSO4- + 3H+ + 2e- → SO2 + 2H2O E= +0.172 V
Only two half reactions involving H3PO3
H3PO3 + 2H+ + 2e-→ H3PO2 + H2O E= (0.499) V
H3PO4 + 2H+ + 2e- → H3PO3 + H2O E= (0.276) V
Which is the most negative?
OR
OFB Chapter 13 263/23/2004
13-4 Concentration Effects and the Nernst Equation
• For Electrochemical Cells which are NOT at standard states (1M or 1 atm)
• From Chapter 11
– Where Q is the reaction quotient• From Chapter 13
Can substitute -n F ∆E° for ∆G° and -n F ∆E for ∆G, and rearranging, becomes the Nernst Equation
OFB Chapter 13 273/23/2004
If Temp. at 25C and
R = 8.3 JK-1mol-1 and
F = 96,485 C mol-1 and use
J/C = V,
then the equation becomes
Nernst Equation
Nernst Equation
OFB Chapter 13 283/23/2004
Example 13-6
The cell
Zn | Zn 2+ || MnO4- | Mn2+ | Pt
(the same cell used in Example 3-3) is setup at 298.15 K with the following non-standard concentrations: [H+] = 0.010M (i.e., acidic Ph=2), [MnO4
-] = 0.12M, [Mn2+] = 0.0010M, and [Zn2+] = 0.015M. Calculate the cell voltage.
Anode (oxidation) | cathode (reduction)
This is a Nernst equation problem
1. Balanced the two half reactions
2. Determined overall # electrons transferred
3. Calculate ∆E° for standard conditions (1M)
4. Using the Nernst equation, use ∆E° andsubstitute the non-standard concentrations and solve.
OFB Chapter 13 293/23/2004
This is a Nernst equation problem
1. Balanced the two half reactions and determined overall # electrons transferred
2. Calculate ∆E° for standard conditions (1M)
3. Using the Nernst equation, use ∆E° andsubstitute the non-standard concentrations and solve.
OFB Chapter 13 303/23/2004
1. Balanced the two half reactions and determine # electrons transferred
Cathode (Reduction)
MnO4-1 → Mn 2+
Zn | Zn 2+ || MnO4- | Mn2+ | Pt
Anode (oxidation) | cathode (reduction)
Anode (Oxidation)
Zn → Zn 2+
OFB Chapter 13 313/23/2004
Overall reaction is
2 MnO4-1 + 5 Zn + 16H+
→ 2Mn 2+ + 5 Zn 2+ + 8 H2O
This is a Nernst equation problem
1. Balanced the two half reactions and determined overall # electrons transferred
2. Calculate ∆E° for standard conditions (1M)
Next calculate ∆E° (look-up in tables)
OFB Chapter 13 323/23/2004
This is a Nernst equation problem
1. Balanced the two half reactions and determined overall # electrons transferred
2. Calculate ∆E° for standard conditions (1M)
3. Using the Nernst equation, use ∆E° andsubstitute the non-standard concentrations and solve.
Overall reaction is
2 MnO4-1 + 5 Zn + 16H+
→ 2Mn 2+ + 5 Zn 2+ + 8 H2O
OFB Chapter 13 333/23/2004
Qn
V.∆∆10
log05920 EE −= °
1621
4
5222
][][][][
+−
++
=HMnO
ZnMnQ
1621
4
5222
10 ][][][][log
100592025.2E
+−
++
−=HMnO
ZnMnV.V∆
162
5223
10 ]01.0[]12.0[]105.1[]10[log
100592025.2E
−−
−=xV.V∆
OFB Chapter 13 343/23/2004 voltage) (Cell Volts 2.16 E
)103.5(log10
0592025.2E 18
10
=
−=
∆
xV.V∆
Review Example 13-6
The cell
Zn | Zn 2+ || MnO4- | Mn2+ | Pt
(the same cell used in Example 3-3) is setup at 298.15 K with the following non-standard concentrations: [H+] = 0.010M (i.e., acidic Ph=2), [MnO4-] = 0.12M, [Mn2+] = 0.0010M, and [Zn2+] = 0.015M. Calculate the cell voltage.
Qn
V.∆∆10
log05920EE −°=
OFB Chapter 13 353/23/2004
Thus far we have used the Nernst equation to calculate the potential difference in a cell.
If given a voltage for a cell, we can calculate an [X] or unknown concentration given the concentrations of the other species
Qn
V.∆∆10
log05920EE −°=
Rearrange the Nernst Equation
OFB Chapter 13 363/23/2004
]EE[05920
log10
∆∆V.
nQ −°=
Q is the reaction quotient
e.g. for the general reaction,
a A + b B → c C + d D
ba
dc
BADCQ
][][][][
=
OFB Chapter 13 373/23/2004
13-5 Equilibrium Constants from Electrochemistry
Previously
If a reaction is at equilibrium, after rearrangement this becomes
°
= E
RTnK ∆
Fln
We can now use cell reactions to find Equilibrium constant for Redox reactions, which are difficult to determine directly.
RT
r∆GlnK
°−= ∆Gr° = − n F ∆E°
OFB Chapter 13 383/23/2004
°∆
= EFln
RTnK
If Temp. at 25C and
R = 8.3 JK-1mol-1 and
F = 96,485 C mol-1 and use
J/C = V,
then the equation becomes
°
= ∆E
V.nK
0592010log
OFB Chapter 13 393/23/2004
Example 13-8
Calculate the equilibrium constant of the Redox reaction
2 MnO4-1 + 5 Zn + 16H+
→ 2Mn 2+ + 5 Zn 2+ + 8 H2O
At 25C using the standard potential difference established in Example 13-3
The cell
Zn | Zn 2+ || MnO4- | Mn2+ | Pt
Anode (oxidation) | cathode (reduction)
∆E°= 2.27 volts
N = 10 electrons
OFB Chapter 13 403/23/2004
An enormous equilibrium constant, which means this reaction equilibrium lies to the right.
2 MnO4-1 + 5 Zn + 16H+
→ 2Mn 2+ + 5 Zn 2+ + 8 H2O
Permanganate is an extremely strong oxidizing agent and Zinc is an extremely strong reducing agent
OFB Chapter 13 413/23/2004
2 MnO4-1 + 5 Zn + 16H+
→ 2Mn 2+ + 5 Zn 2+ + 8 H2O
Permanganate is an extremely strong oxidizing agent and Zinc is an extremely strong reducing agent
110
][][][][ 380
16214
5222== +−
++
HMnOZnMnK
OFB Chapter 13 423/23/2004
• Electrolytic Cells can also be used to measure– Acidity constants– Basicity constants– Solubility product constants
AgCl (s) excess → Ag +1 + Cl -1
To determine Ksp, just find the [Ag+1] in an electrolytic cell containing a known amount of [Cl-1]
OFB Chapter 13 433/23/2004
• Example 13-9A galvanic cell consists of a standard hydrogen half-cell (with platinum electrode) operating as the anode, and a silver half-cell:
Pt|H2 (1 atm)|H+ (1M)||Ag+|AgThe Ag+ (aq) in the cathode compartment is in equilibrium with some solid AgCl and Cl- (aq); the concentration of the Cl- (aq) is 0.00100 M. The measured cell voltage is ∆E=0.398V. Calculate the silver-ion concentration in the cell and the Ksp of silver chloride at 25°C
OFB Chapter 13 463/23/2004
Overall reaction
2 Ag + + H2 → 2 H + + 2 Ag
7
132
132
22
2
22
106.1][
104][
1
1041][
1]1[
][][][
−+
+
+
+
+
=
==
==
=
xAg
xAg
Q
xAg
Q
PAgAgHQ
H[H+] Given
[Cl-] Given
Ag ppt
Calculated
Answer10
37
106.1
]101][106.1[
]][[
−
−−
−+
=
=
=∴
xK
xxK
ClAgK
AgClsp
AgClsp
AgClsp
Pt|H2 (1 atm)|H+ (1M)||Ag+|Ag
PH2Given
OFB Chapter 13 473/23/2004
Chapter 13 Summary
∆G = - Q ∆E = -n F ∆E(at constant T,P)
∆G = change in Gibbs EnergyQ = charge∆E = potential differenceN = # electrons transferredF = Faraday constant
][05920
log10 EE
∆∆V.
nQ −°=
∆E° = E° (cathode) - E° (anode)
SuppliesIncreaseReducing AgentPicksDecreaseOxidizing AgentGainDecreaseReductionLossIncreaseOxidation∆Electron∆ OxidizationTerm