Redox Reactions and Electrochemistry
• One of the most important areas of Applied Thermodynamics
• Study of the relationship between chemical change and electrical work. This
relationship is investigated through the use of electrochemical cells
• The electrochemical cells incorporate a redox reaction to produce or to utilize
electrical energy
Electron-transfer reactions: Cu (s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag (s)
Redox Reactions
Cu (s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag (s)
Cu (s): from 0 to +2 oxidized reducing agent
Ag (s): from +1 to 0 reduced oxidizing agent
• Oxidation (electron loss) is always accompanied by reduction (electron gain)
• Oxidizing agent is reduced the reducing agent is oxidized
• The total number of e − gained by the oxidizing agent is always equal to the total
number lost by the reducing agent
Half-reaction method for balancing redox reactions
• It divides the overall redox reaction into oxidation and reduction half-rxns
• Each half-rxn is balanced for mass (atoms) and charge
This method is very useful because
a) it separates the oxidation and reduction steps, which reflect their actual physical
separation in electrochemical cells
b) it is readily applied to redox reactions that take place in acidic or basic solutions
(common in cells)
Note: Review Oxidation Numbers
Example Balance the reaction in acidic solution
CuS (s) + NO 3− (aq) → Cu2+ (aq) + SO 24− (aq) + NO (g)
1
1. Write two half-rxns:
CuS (s) → Cu2+ (aq) + SO 24− (aq)
NO 3− (aq) → NO (g)
2. Balance elements other than H and O:
Balanced
3. Balance oxygen by inserting H2O
CuS (s) + 4H2O (l) → Cu2+ (aq) + SO 24− (aq)
NO 3− (aq) → NO (g) + 2H2O (l)
4. Balance hydrogen by inserting H+
CuS (s) + 4H2O (l) → Cu2+ (aq) + SO 24− (aq) + 8H+ (aq)
NO 3− (aq) + 4H+ (aq) → NO (g) + 2H2O (l)
5. Balance charge by inserting e −
CuS (s) + 4H2O (l) → Cu2+ (aq) + SO 24− (aq) + 8H+ (aq) + 8e − (oxidation)
NO 3− (aq) + 4H+ (aq) + 3e − → NO (g) + 2H2O (l) (reduction)
6. Make the number of e − in the two half-rxns the same
3CuS (s) + 12H2O (l) → 3Cu2+ (aq) + 3SO 24− (aq) + 24H+ (aq) + 24e −
8NO 3− (aq) + 32H+ (aq) + 24e − → 8NO (g) + 16H2O (l)
7. Addition of the two half-rxns
3CuS (s) + 12H2O (l) + 8NO 3− (aq) + 32H+ (aq) + 24e − → 3Cu2+ (aq) + 3SO 24− (aq) +
24H+ (aq) + 24e − + 8NO (g) + 16H2O (l)
3CuS (s) + 8NO 3− (aq) + 8H+ (aq) → 3Cu2+ (aq) + 8NO (g) + 3SO 24− (aq) + 4H2O (l)
Check: Mass balance Cu: 3 3, S: 3 3, N: 8 8, H: 8 8, O: 24 24
Charge balance -8+8=0 +6-6=0
Example Balance the reaction in basic solution
Ag (s) + HS − (aq) + CrO 24− (aq) → Ag2S (s) + Cr(OH)3 (s)
1. Write the two half-rxns
Ag (s) + HS − (aq) → Ag2S (s)
2
CrO 24− (aq) → Cr(OH)3 (s)
2. Balance the elements other than H and O
2Ag (s) + HS − (aq) → Ag2S (s)
CrO 24− (aq) → Cr(OH)3 (s)
3. Balance O by inserting H2O
2Ag (s) + HS − (aq) → Ag2S (s)
CrO 24− (aq) → Cr(OH)3 (s) + H2O (l)
4. Balance H by inserting H2O on the side that is deficient in hydrogen and an equal
amount of OH − on the other side (basic solution)
2Ag (s) + HS − (aq) + OH − (aq) → Ag2S (s) + H2O (l)
CrO 24− (aq) + 5H2O (l) → Cr(OH)3 (s) + H2O (l) + 5OH − (aq)
5. Balance charge by inserting e −
2Ag (s) + HS − (aq) + OH − (aq) → Ag2S (s) + H2O (l) +2e − (oxidation)
CrO 24− (aq) + 5H2O (l) +3e − → Cr(OH)3 (s) + H2O (l) + 5OH − (aq) (reduction)
6. Make the number of e − in the two half-rxns the same
6Ag (s) + 3HS − (aq) + 3OH − (aq) → 3Ag2S (s) + 3H2O (l) +6e −
2CrO 24− (aq) + 10H2O (l) +6e − → 2Cr(OH)3 (s) + 2H2O (l) + 10OH − (aq)
7. Addition of the two half-rxns
6Ag (s) + 3HS − (aq) + 3OH − (aq) + 2CrO 24− (aq) + 10H2O (l) +6e − → 3Ag2S (s) +6e −
+2Cr(OH)3 (s) + 2H2O (l) + 2OH − (aq)
6Ag (s) + 3HS − (aq) + 2CrO 24− (aq) + 5H2O (l) → 3Ag2S (s) + 2Cr(OH)3 (s) +
7OH − (aq)
Check: Mass balance Ag: 6 6, S: 3 3, Cr: 2 2, H: 13 13, O: 13 13
Charge balance -7 -7
Disproportionation Reactions
In this kind of reactions the same chemical species is both oxidized and reduced; it
reacts with itself. Balancing this type of reactions is straightforward by the half-rxn
method once it is realized that the same species may appear on the left in both half –rxns.
Example: Cl2 (aq) → ClO 3− (aq) + Cl − (aq) (unbalanced)
3
Step 1 gives:
Cl2 (aq) → ClO 3− (aq)
Cl2 (aq) → Cl − (aq)
Proceeding with steps 2 to 5 yields
Cl2 (aq) + 6H2O (l) → 2ClO 3− (aq) + 12H+ (aq) + 10e − (oxidation)
Cl2 (aq) +2e − → 2Cl − (aq) (reduction)
Make the number of e − in the two half-rxns the same
6Cl2 (aq) + 6H2O (l) → 2ClO 3− (aq) + 12H+ (aq) + 10Cl − (aq)
Divide the coefficients by 2:
3Cl2 (aq) + 3H2O (l) → ClO 3− (aq) + 6H+ (aq) + 5Cl − (aq)
Electrochemical Cells
Electric charge (C) coulombs
Charge on e − 1.602 x 10-19 C
1 mol of e − : (1.602 x 10-19 C)(6.022 x 1023 mol-1) = 96, 485 C
Faraday’s constant (F)
Relation between charge and mole: q = n e F
Electric current: the quantity of charge flowing each second through a circuit. The unit
is called Ampère (A): charge of 1C per second flowing past a point in a circuit.
The difference in electric potential, E, between two points is the work needed (or that
can be done) when moving an electric charge from one point to another. It is measured
in volts (V)
Volt = Joule/Coulomb (J/C)
Ohm’s Law: I =
E
R
(R is called resistance, units ohms (Ω))
1 Ω is the current of 1 A that flows through a circuit with a potential difference of 1 volt
if the resistance of the circuit is 1 Ω
4
Anode: oxidation
Cathode: reduction
E cell = E cathode − E anode
Line notation: | (phase boundary)
|| (salt bridge)
Cd(s)|Cd(NO3)2(aq)||AgNO3(aq)|Ag(s)
Galvanic (voltaic) cell: Electrons flow
spontaneously from the negative to the
positive electrode. Ecell > 0
Conversion of chemical potential
energy into electrical energy that can
perform work
Electrical energy is proportional to Ecell
An electrolytic cell uses electrical energy to carry out a chemical rxn that would
otherwise not occur. (non-spontaneous cells)
Difference between galvanic and electrolytic cells
5
Fundamentals of Electrolysis
Michael Faraday, the father of electrochemistry
Faraday’s Laws
• The quantities of substances produced and consumed at the electrodes are directly
proportional to the amount of electric charge passing through the cell
• When a given amount of electric charge passes through a cell, the quantity of a
substance produced or consumed at an electrode is proportional to its molecular
mass divided by the number of moles of electrons required to produce or consume
one mole of the substance.
If a current I flows for time t, the charge q passing any point in the circuit is:
q=It
q Coulombs, I Ampères, t seconds
coulombs
It
=
coulombs/mole F
If a reaction requires n electrons per molecule, the quantity reacting in time t is:
It
Moles reacted =
ne − F
Example
If a current of 0.17A flows for 16 min through the cell in the
adjacent figure how many grams of Cu will be deposited?
The moles of e − are:
From the diagram we conclude that:
Cathode: Cu2+ + 2e −  Cu (s)
Anode:
H2O  ½ O2 (g) + 2H+ + 2e −
Net rxn: H2O + Cu2+  Cu (s) + ½ O2 (g) + 2H+
moles of e − :
It (0.17C/s)(16min)(60s/min)
=
=1.69x10-3 mol
F
96,485C/mol
The cathode half-reaction requires 2e − for each Cu(s)
deposited.
Moles of Cu(s) = ½ (moles of e − ) = 8.45 x 10-4 mol, or (8.45 x
10-4 mol)(63.546 g mol-1) = 0.054 g.
You may use also this equation for mass of solid deposited or mass of gas evolved:
1 Atomic Mass
m=
It
F
ne−
6
Example
Draw a diagram, show balanced equations, and write the notation for a voltaic cell that
consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an
Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the
Cr electrode is negative relative to the Ag electrode.
Identify the oxidation and reduction reactions and write each half-reaction. Associate
the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction).
Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)
Molten salts: refer to a salt that is in the liquid phase that is normally a solid at standard
temperature and pressure. A salt that is normally a liquid at STP is usually called a room
temperature ionic liquid, although technically molten salts are a class of ionic liquids.
Molten salts have a variety of uses. Molten chloride salt mixtures are commonly used as
baths for various alloy heat treatments such as annealing of steel. Cyanide and chloride
salt mixtures are used for surface modification of alloys, fluoride, chloride, and nitrate
can also be used as heat transfer fluids as well as for thermal storage.
7
Example: Electrolysis of molten MgCl2
Mg2+ (melt) + 2Cl − (melt) → Mg (l) + Cl2 (g)
Half-rxns: 2Cl − (melt) → Cl2 (g) + 2e − (oxidation)
Mg2+ (melt) + 2e − → Mg (l) (reduction)
Cell: graphite | Cl2 (g) | Cl − (melt), Mg2+ (melt) | Mg (l) | steel
The Gibbs Free Energy and Cell Voltage
When an amount of charge, Q, moves through a potential difference, ∆E
w = - Q ∆E
↑ b/c work done by the system
∆E > 0 for galvanic (voltaic) cells
Recall, G = H – TS = E + PV – TS
For constant T and P: ∆G = ∆E + P∆V - T∆S (usual case in electrochemical cells)
But ∆E = q + w
Total work = P-V work + electrical work (w elect) = - P∆V + w elect
∆E = q + w = q + w elect - P∆V
∆G = q + w elect - P∆V + P∆V - T∆S = = q + w elect - T∆S
If the condition of reversibility is imposed upon the galvanic cells, q = q rev = T∆S
Therefore, ∆G = q rev + w elect - T∆S = T∆S + w elect - T∆S = w elect
And w elect = - Q ∆E
If n moles of e − (nF coulombs of charge) pass through the external circuit of the
galvanic cell when it is operated reversibly and if ∆E is the reversible voltage, then
∆G rxn = - Q ∆E = - nF∆E (why ∆G rxn ? we’ll prove it in a bit!)
Electrical work is produced when ∆G rxn < 0
8
Standard States and Cell Voltages
∆G orxn = - n F∆Eo, ∆Eo (standard cell voltage): is the cell voltage in a galvanic cell in
which reactants and products are in their standard states. (Gases at 1 bar, solutes in 1 M,
metals in their pure stable states and at a specified temperature)
Example
A 6.0 Volt battery delivers a steady current of 1.25 A for a period of 1.50 hours.
Calculate the total charge in coulombs that passes through the circuit.
Q = It = (1.25 C s-1)(1.5 hours)(3600 s hour-1) = 6750 C
w elect = - Q ∆E = - (6750 C)(6 V) = - 4.05 x 104 J (work done on the battery)
- (- 4.05 x 104 J) = + 4.05 x 104 J work done by the battery.
Half-Cell Potentials
• It is a very long and tedious job to tabulate all the conceivable galvanic cells
• We can avoid the previous job by tabulating the half-cell reduction potentials, Eo
• These Eo values express the intrinsic tendency of a reduction half-reaction to
occur when the reactants and products are at standard states.
o
o
∆E o = E cathode
− E anode
How Eo’s are measured?
Line notation of cell:
S.H.E. || Ag+ (aq,1M) | Ag(s)
Oxidation occurs in the SHE half-cell
and this electrode is the anode:
H2 (g 1 bar) → 2H+ (aq,1M) + 2e −
2H+ (aq,1M) + 2e − → H2 (g,1 bar)
Eo (H2 / H+, oxid. = Eo (H+ / H2 , red.
≡ 0.00 V
9
Consider Cu2+ (aq,1 M)|Cu (s) half-cell connected to SHE. We measure ∆E o = 0.34 V
Then we assign Eo for Cu2+ (aq,1 M) + 2e − → Cu (s)
Now consider Zn2+ (aq,1 M)|Zn (s) half-cell connected to SHE. We measure
∆E o = - 0.76 V. Then we assign Eo = -.0.76V for Zn2+ (aq,1 M) + 2e − → Zn (s)
Now we build the cell: Zn2+ (aq,1 M)|Zn (s) || Cu2+ (aq,1 M)|Cu (s) for which
o
o
∆E o = Ecathode
− Eanode
= 0.34 − ( −0.76) = 1.10V (>0) - Cell is galvanic
Half-cell potentials are intensive properties, namely independent of the amount of the
reacting species.
1. All values are relative to SHE ( = reference electrode)
2. Half-reactions are written as reductions (only reactants are oxidizing agents and
only products are the reducing agents)
3. The more positive the Eo the more readily the reaction occurs
4. Half-reactions are shown with ℑ b/c each can occur as reduction or oxidation
5. The half-cell that is listed higher at the table acts as the cathode
Writing spontaneous redox reactions
a) By convention, electrode potentials are written as reductions
b) When pairing two half-cells, you must reverse one reduction half-cell to produce an
oxidation half-cell. Reverse the sign of the potential.
10
c) The reduction half-cell potential and the oxidation half-cell potential are added to
obtain the E ocell .
d) When writing a spontaneous redox reaction, the left side (reactants) must contain the
stronger oxidizing and reducing agents.
Example
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
stronger
reducing
agent
stronger
oxidizing
agent
weaker
oxidizing
agent
weaker
reducing
agent
Example
Will Ag+ oxidize Zn (s) or will Zn2+ oxidize Ag (s)
Example
A voltaic cell houses the reaction between aqueous bromine and zinc metal:
o
= 1.83 V
Br2+ (aq) + Zn (s) → Zn2+ (aq) + 2Br - (aq) Ecell
o
o
Calculate Ebro
min e given E zinc = - 0.76 V.
o
= 1.83 V > 0 (reaction spontaneous) as written. Zn is being
The cell is voltaic Ecell
oxidized and therefore it is the anode.
o
o
o
o
o
o
o
Ecell
= Ecathode
− Eanode
⇒ Ecathode
= Ecell
+ Eanode
= 1.83 + ( −0.76) = 1.07 V = Ebro
min e
Example
A standard (Pt | MnO 4− , H+, Mn2+) half-cell consists of an inert electrode in contact with
a solution containing MnO 4− (aq) , H+ (aq) and Mn2+ (aq) ions in standard states. Such a
half-reaction is assembled and connected to a standard Zn2+ (aq,1 M)|Zn (s) half-cell/
a) Calculate ∆E of the cell at 25oC.
In the Handbook of Physics and Chemistry we find:
(Rxn 1): MnO 4− (aq) + 8H+ (aq) + 5e − → Mn2+ (aq) + 4H2O (l) Eo = 1.507V
(Rxn 2): Zn2+ (aq) + 2e − → Zn (s) Eo = - 0.762V
Rxn 1 is the cathode because it lies above Eo (Zn2+/Zn), So
o
o
∆E o = Ecathode
− Eanode
= 1.507 − ( −0.762) = +2.269 V
b) Write a balanced equation for the overall reaction
11
We multiply (Rxn 1) by 2 and (Rxn 2) by 5:
2MnO 4− (aq) + 16H+ (aq) + 5Zn (s) → 2Mn2+ (aq) + 5Zn2+ (aq) + 8H2O (l)
Number of e − involved: 10
Now check:
∆G orxn = [2∆G of (Mn2+ (aq)) +5∆G of (Zn2+ (aq))+ 8∆G of (H2O (l)) ] - [2∆G of (MnO 4− (aq))
16∆G of (H+ (aq)) +5∆G of (Zn (s)) ] = - 2194.5 kJ
o
∆E=
o
∆Grxn
-2194.5x103 Jmol-1
=2.27JC-1 =2.27V Very good match!!!
=
-1
(10)(96,485Cmol )
ne F
Disproportionation
• A single chemical species is both oxidized and reduced
• This species must be able to give up electrons and accept electrons and in
addition the half-reaction in which it is reduced must lie higher in the table than
the half-reaction in which it is oxidized
• If this is the case then it drives the second reaction to go in reverse and
disproportionation occurs spontaneously
Example
(1): Cu+ (aq) + e − → Cu (s) Eo = 0.521 V
(2): Cu2+ (aq) + e − → Cu+ (aq) Eo = 0.153V
(1) is reduction and drives (2) as oxidation
2 Cu+ (aq) → Cu2+ (aq) + Cu (s) E ocell = 0.521 – 0.153 = 0.368 V (>0)
Which means E orxn > 0 and ∆G orxn < 0 spontaneous. Yes it disproportionates.
Example
Decide whether Fe2+ (aq) in its standard state at 25oC is stable w/r/t
disproportionation.
In Appendix we find:
(1): Fe3+ (aq) + e − → Fe2+ (aq) Eo = 0.771V
(2): Fe2+ (aq) + 2e − → Fe (s) Eo = - 0.447V
12
In order for Fe2+ (aq) to disproportionate reaction (1) would have to be driven
backwards (oxidation): 3Fe2+ → Fe (s) + 2Fe3+ (aq)
And Eo = -0.447 – (+0.771) = -1.218 V (< 0) and ∆G orxn > 0 (non-spontaneous) and
therefore stable against disproportionation.
Oxygen as an oxidizing agent
O2 (g) + 4H+ (aq) + 4e − → 2H2O (l) Eo = 1.229V good oxidizing agent!
O3 (g) + 2H+ (aq) + 2e − → O2 (g) + H2O (l) Eo = 2.067V even better oxidizing agent!
H2O2 (aq) + 2H+ (aq) + 2e − → 2H2O (l) Eo = 1.776V (bleach + germicider)
O2 (g) +2H+ (aq) + 2e − → H2O2 (aq) Eo = 0.695V
H2O2 as a reducing agent can reduce only chemical species with reduction potentials
greater than 0.695 V.
Relative Reactivities of Metals
• Metals that can displace hydrogen from acid
Fe (s) + 2H+ (aq) → H2 (g) + Fe2+ (aq) Why does this rxn go?
Fe (s) → Fe2+ (aq) + 2e − Eo = - 0.44V and 2H+ (aq) + 2e − → H2 (g) Eo = 0.0 (SHE)
E ocell = 0.00 – ( - 0.44) = + 0.44 V (> 0) rxn will go!
• Metals that cannot displace hydrogen from acid
2Ag (s) + 2H+ (aq) → 2Ag+ (aq) + H2 (g) Eo = 0.80V
2 x [Ag (s) → Ag+ (aq) + e − ] and 2H+ (aq) + 2e − → H2 (g) Eo = 0.0 (SHE)
E ocell = 0.00 – (0.80) = - 0.80 V (< 0) rxn will not go!
• Metals that can displace hydrogen from water
2H2O (l) + 2e − → H2 (g) + 2OH − (aq) Eo = - 0.42 V
However, [OH − ] = 10-7 non-standard
Metals active enough to reduce H2O (l) lie below ( – 0.42 V)
2Na (s) + 2H2O (l) → 2Na+ (aq) + H2 (g) + 2OH − (aq)
2 x [Na (s) → Na+ (aq) + e − ] Eo = -2.71 V
2H2O (l) + 2e − → H2 (g) + 2OH − (aq) Eo = -0.42V,
E ocell = -0.42 – (-2.71) = 2.29 V Yes, Na can!
• Metals that displace other metals in solutes
Zn (s) + Fe2+ (aq) → Zn2+ (aq) + Fe (s)
13
Zn (s) → Zn2+ (aq) + 2e − Eo = - 0.76 V
Fe2+ (aq) + 2e − → Fe (s) Eo = - 0.44 V
E ocell = -0.44 – (- 0.76) = 0.32 V
Concentration Effects and the Nernst Equation
∆G = ∆G o + RT ln Q
∆G o = − nFE o
∆G = − nFE
∆G = ∆G o + RT ln Q
− nFE = − nFE o + RT ln Q
Divide both sides – nF
− nFE − nFE o
RT
ln Q
=
+
− nF
− nF
− nF
RT
E = Eo −
ln Q Nernst Equation (expresses the net driving force for a reaction)
nF
Eo: standard reduction potential
R: 8.314 J K-1 mol-1 = 8.314 V C K-1 mol-1
F: Faraday’s constant = 96,485 C mol-1
T: absolute temperature (K)
n: number of electrons in the half-reaction
Q: reaction quotient
RT
Let us evaluate
at 25oC (298.15K)
nF
RT
nF
(8.314JK -1mol-1 )(298.15K) 0.0257
=
n(96,485Cmol-1 )
n
o
And the Nernst Equation at 298.15K becomes, E = E −
o
Recall that ln x = 2.303 log10 x and E = E −
0.0257
ln Q
n
0.05916
log10 Q
n
Recall, at equilibrium ∆G = 0 ∴ ∆G o = − RT ln K ⇒ − nFE o = − RT ln K
o
And E =
RT
ln K
nF
Let’s see in depth!
E = Eo −
RT
ln Q
nF
When Q < 1 and thus [reactant] > [product], ln Q < 0, so E > E ocell
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When Q = 1 and thus [reactant] = [product], ln Q = 0, so E = E ocell
When Q >1 and thus [reactant] < [product], ln Q > 0, so E < E ocell
The signs of ∆Go and E ocell determine the direction of the reaction at standard conditions.
Example
Suppose that the Zn |Zn2+(aq) || MnO 4− (aq) | Mn2+(aq) |Pt cell is operated at pH 2.00
with [MnO 4− ]= 0.12 M, [Mn2+|] = 0.0010 M and [Zn2+] = 0.015 M. Calculate the cell
voltage, ∆E , at 25 oC.
The reaction of this cell is:
2MnO 4− (aq) + 16H+ (aq) + 5Zn (s) → 2Mn2+ (aq) + 5Zn2+ (aq) + 8H2O (l) , ∆Eo = 2.27V
0.0257
0.0257 [Mn 2+ ]2 [Zn 2+ ]5
0.0257 (0.0010) 2 (0.015)5
ln Q= 2.27 −
ln
ln
∆E= ∆E −
= 2.27 −
= 2.14V
10
10
[MnO-4 ]2 [H + ]16
10
(0.12) 2 (0.010)16
o
Example
An electrochemical cell is set up at 25oC. One half-cell consists of a zinc anode
immersed in a 1.0 M solution of Zn(NO3)2. The second half-cell consists of a platinum
cathode that has gaseous hydrogen bubbling over it at a pressure of 1.00 atm in a
solution of unknown hydrogen-ion concentration. The observed cell voltage is 0.472 V.
(a) Calculate the reaction quotient, Q
15
Anode: Zn (s) →Zn2+ (aq) + 2e − Eo = - 0.762V
Cathode: 2H+ (aq) + 2e − → H2 (g) Eo = 0.00 V (SHE)
∆Eo = 0.00 – (-0.762) = 0.762 V
The cell is not at the standard state
∆E = ∆E o −
RT
nF
2
ln Q ⇒ ln Q =
( ∆E o − ∆E ) =
(0.762 − 0.427) = 22.48
nF
RT
0.0257
Q = e 22.48 ≈ 5.8 x10 9
(b) Calculate the hydrogen-ion concentration in the second half-cell
[Zn 2+ ]PH 2 (1.00)(1.00)
=
Q =
=5.8x109 Þ[H + ]=1.3x10-5 M
+ 2
+ 2
[H ]
[H ]
Electrolytic Cells
In galvanic (voltaic) cells electrons are generated at the anode (-) and they are consumed
st the cathode (+)
In an electrolytic cell, electrons come from an external power source, which supplies to
the cathode and removes them from the anode. Cathode (-), Anode (+)
The tin-copper reaction as the basis of a voltaic and an electrolytic cell
At standard conditions the
spontaneous reaction
between Sn and Cu2+
generates 0.48 V (voltaic)
If more than 0.48 V
supplied the same
apparatus and components
become an electrolytic
cell and the nonsponatneous reaction
16
between Cu and Sn2+ occurs.
Concentration Cells
In these cells the half-reactions are the same but the concentrations are different.
A concentration cell based on the Cu/Cu2+ half-reaction
The easiest is pH.
We construct a cell with cathode to be SHE and the unknown has the same apparatus
dipping into an unknown [H+] solution.
Oxidation: H2 (g 1 atm) → 2H+ (aq, unknown) + 2e −
Reduction: 2H+ (aq, 1M) + 2e − → H2 (g 1 atm)
17
Overall: 2H+ (aq, 1M) → 2H+ (aq, unknown)
Ecell
[H + ]2umknown
0.05916
0.05916
log + 2
0.00 −
log[H + ]2 =-0.05916log[H + ]
=
E −
=
2
[H ]standard
2
o
cell
By measuring the cell potential we can get [H+] and consequently pH
pH Meter
Example
A concentration cell consists of two Ag/Ag+ half-cells. In half-cell A, electrode A dips
into 0.010 M AgNO3; in half-cell B, electrode B dips into 4.0 x 10-4 M AgNO3. What is
the cell potential at 298.15 K? Which electrode has a positive charge?
Ag+(aq, 0.010 M) [half-cell A] → Ag+(aq 4.0 x 10-4 M) [half-cell B]
[Ag + ]B
0.05916
0.05916
4.00x10-4
o
Ecell =
Ecell
−
=
log
0.00log
=0.0828V
1
[Ag + ]A
1
0.010
Reduction occurs at the cathode electrode A, so electrode A has a positive charge.
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