Chapter21 PDF
Chapter21 PDF
Chapter21 PDF
reduction separately
– Electrochemistry deals with the relationship
between chemical change and electricity Example: Ca(s) + Cl2(g) → CaCl2(s)
→CaCl2(s) consists of Ca2+ and Cl- ions
– Electrochemical cells (two types)
Ca(s) → Ca2+(s) + 2e- (loss of 2e-, oxidation)
• Galvanic cells – use a spontaneous (∆G < 0) reaction + Cl (g) + 2e- → 2Cl-(s) (gain of 2e-, reduction)
to produce electricity (batteries) 2
• Electrolytic cells – use a source of electricity to drive Ca(s) + Cl2(g) + 2e- → Ca2+(s) + 2e- + 2Cl-(s)
a non-spontaneous (∆G > 0) reaction (electrolysis)
→Adding the half-reactions gives the overall reaction
21.1 Redox Half-Reactions →Ca is oxidized (Ca is the reducing agent)
• Redox reactions involve e- transfer →Cl2 is reduced (Cl2 is the oxidizing agent)
¾Generalized expressions for half reactions:
– Oxidation – loss of e- (oxidation state ↑) ¾Red → Ox + ne- or Ox + ne- → Red
– Reduction – gain of e- (oxidation state ↓) ¾Ox/Red form a redox couple (Ex: Ca2+/Ca; Cl2/Cl-)
1
3. Multiply the 1st half-reaction by 2 and the 2nd by 3 21.2 Galvanic (Voltaic) Cells
to get 6e- in both half-reactions
• Produce electricity from a spontaneous
2CrO2- + 4H2O → 2CrO42- + 8H+ + 6e- chemical reaction
+
3BrO4- + 6H+ + 6e- → 3BrO3- + 3H2O Example: Zn metal reacts spontaneously with
4. Add the half-reactions Cu2+ solutions to yield metallic Cu and Zn2+ ions
2CrO2- + 4H2O + 3BrO4- + 6H+ + 6e- → Zn(s) + Cu2+ → Zn2+ + Cu(s) (SO42- counter ions)
→ 2CrO42- + 8H+ + 6e- + 3BrO3- + 3H2O →The two half-reactions are:
⇒ 2CrO2- + 3BrO4- + H2O → 2CrO42- + 3BrO3- + 2H+ Zn(s) → Zn2+ + 2e- (oxidation)
Cu2+ + 2e- → Cu(s) (reduction)
5. Add 2OH- on both sides of the equation
2H2O →The two half-reactions can be physically separated
2CrO2- + 3BrO4- + H2O + 2OH- → by placing them in separate containers (half-cells)
→ 2CrO4 + 3BrO3 + 2H+ + 2OH-
2- -
→Half-cells → where the half-reactions occur
2CrO2- + 3BrO4- + 2OH- → 2CrO42- + 3BrO3- + H2O →Anode half-cell → where oxidation occurs
→Cathode half-cell → where reduction occurs
2
¾Electrodes involving gases – a gas is bubbled Example: A combination of the Zn(s)Zn2+ and
over an inert electrode Fe3+, Fe2+Pt(s) half-cells leads to:
Example: H2 gas over Pt electrode
H2(g) → 2H+ + 2e- (as oxidation)
Notation: Pt(s)H2(g)H+
• Cell notation
– The anode half-cell is written on the left of the
cathode half-cell Zn(s) → Zn2+ + 2e- (anode, oxidation)
– The electrodes appear on the far left (anode) and +
Fe3+ + e- → Fe2+ (×2) (cathode, reduction)
far right (cathode) of the notation
Zn(s) + 2Fe3+ → Zn2+ + 2Fe2+
– Salt bridges are represented by double vertical
lines ⇒ Zn(s)Zn2+ || Fe3+, Fe2+Pt(s)
Example: A combination of the Pt(s)H2(g)H+ Example: Write the cell reaction and the cell
and Cl-AgCl(s)Ag(s) half-cells leads to: notation for a cell consisting of a graphite cathode
Note: The immersed in an acidic solution of MnO4- and Mn2+
reactants in the and a graphite anode immersed in a solution of Sn4+
overall reaction are and Sn2+.
in different phases →Write the half reactions (a list of the most common
(no physical half-reactions is given in Appendix D)
contact) ⇒ no need Sn2+ → Sn4+ + 2e- ×5 (oxidation)
of a salt bridge + MnO - + 8H+ + 5e- → Mn2+ + 4H O(l) ×2 (reduction)
4 2
H2(g) → 2H+ + 2e- (anode, oxidation) 5Sn2+ + 2MnO4- + 16H+ + 10e- → 5Sn4+ + 10e- +
+
AgCl(s) + e- → Ag(s) + Cl- (×2) (cathode, reduction) + 2Mn2+ + 8H2O(l)
2AgCl(s) + H2(g) → 2Ag(s) + 2H+ + 2Cl- →The graphite (C) electrodes are inactive
⇒ Pt(s)H2(g)H+, Cl-AgCl(s)Ag(s) ⇒ C(s)Sn2+, Sn4+ || H+, MnO4-, Mn2+C(s)
3
¾Ecell is measured with a voltmeter • Electrode potentials (E) – characterize the
¾If the (+) terminal of the voltmeter is connected individual electrodes (half-reactions)
to the (+) electrode (cathode), the voltmeter – The cell potential is the difference between the
shows a positive reading electrode potentials of the cathode and anode
¾Ecell characterizes the overall cell reaction Ecell = Ecathode – Eanode
¾If Ecell > 0, the cell reaction is spontaneous
• Standard electrode potentials (Eo) –
¾If Ecell < 0, the cell reaction is non-spontaneous electrode potentials at the standard-state
¾If Ecell = 0, the cell reaction is at equilibrium Eocell = Eocathode – Eoanode
Example: Zn(s)Zn2+(1M) || Cu2+(1M)Cu(s) – Eo values are reported for the half-reaction
+1.10 V written as reduction (standard reduction
potentials) → listed in Appendix D
Zn(s) +Cu2+→ Zn2+ + Cu(s)
Eo cell = 1.10 V > 0 → spontaneous reaction
¾Absolute values for E and Eo can’t be measured – If the unknown electrode is the cathode of the cell
⇒ A reference electrode (half-cell) is needed → Eocell = Eounkn – Eoref
• The potentials of all electrodes are measured relative → Eounkn = Eocell + Eoref = Eocell + 0 = Eocell > 0
to the reference electrode – If the unknown electrode is the anode of the cell
• Standard hydrogen electrode – used as a → Eocell = Eoref – Eounkn
reference electrode → Eoref = 0 V (assumed) → Eounkn = Eoref – Eocell = 0 – Eocell = –Eocell < 0
H+(1M)H2(g, 1atm)Pt(s) Example:
2H+(1M) + 2e- → H2(g, 1atm) Pt(s)H2(g, 1atm)H+(1M), Cl-(1M)AgCl(s)Ag(s)
– To find the potential of any electrode, a cell is H+/H2 → anode
constructed between the unknown electrode and Ag/AgCl → cathode
the reference electrode Eocell = EoAg/AgCl – Eref
= EoAg/AgCl
– The cell potential is directly related to the
unknown electrode potential EoAg/AgCl = +0.22 V
4
Strengths of Oxidizing and Reducing Agents • Electrochemical series – an arrangement of the
• Eo values are always tabulated for reduction redox couples in order of decreasing reduction
potentials (Eo) → Appendix D
Ox + ne- → Red (Eo)
– The most positive Eos are at the top of the table
– Ox is an oxidizing agent; Red is a reducing agent
– The most negative Eos are at the bottom of the table
• Eo is a measure for the tendency of the half-reaction
to undergo reduction ⇒The strongest oxidizing agents (Ox) are at the top
of the table as reactants
⇒Higher (more positive) Eo means ⇒The strongest reducing agents (Red) are at the
– Greater tendency for reduction bottom of the table as products
– Lower tendency for oxidation • Every redox reaction is a sum of two half-reactions,
⇒Higher (more positive) Eo means one occurring as oxidation and another as reduction
– Stronger oxidizing agent (Ox) ← Ox is reduced Red1 → Ox1 + ne- Ox2 + ne- → Red2
– Weaker reducing agent (Red) ← Red is oxidized Red1 + Ox2 → Ox1 + Red2
• In a spontaneous redox reaction, the stronger b) Can Cl2 oxidize H2O to O2 in acidic solution?
oxidizing and reducing agents react to produce the → Cl2/Cl- has higher Eo (Cl2/Cl- is above O2,H+/H2O)
weaker oxidizing and reducing agents ⇒ Cl2 is a stronger oxidizing agent than O2
Stronger Red1 + Stronger Ox2 → Weaker Ox1 + Weaker Red2 ⇒ Cl2 can oxidize H2O to O2 at standard conditions
Example: Given the following half-reactions: c) Write the spontaneous reaction between the Cl2/Cl-
Cl2(g) + 2e- → 2Cl- Eo = +1.36 V and Fe3+/Fe2+ redox couples and calculate its Eocell
Appendix D
O2(g) + 4H+ + 4e- → 2H2O(l) Eo = +1.23 V → Cl2/Cl- has the higher reduction potential (Eo)
Fe + e → Fe
3+ - 2+ Eo = +0.77 V
⇒ Cl2/Cl- undergoes reduction
Fe + 2e → Fe(s)
2+ - Eo = –0.44 V ⇒ Fe3+/Fe2+ undergoes oxidation (reverse equation)
a) Rank the oxidizing and reducing agents by strength Cl2(g) + 2e- → 2Cl- (reduction) Eo = +1.36 V
+
→ Ox agents on the left; Red agents on the right Fe2+ → Fe3+ + e- ×2 (oxidation) Eo = +0.77 V
Oxidizing → (Top) Cl2 > O2 > Fe3+ > Fe2+ (Bottom) Cl2(g) + 2e + 2Fe → 2Cl- + 2Fe3+ + 2e-
- 2+
Reducing → (Bottom) Fe > Fe2+ > H2O > Cl- (Top) Eocell = Eocath – Eoanod = +1.36 – (+0.77) = +0.59 V
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Example: Can Fe and Cu be dissolved in HCl(aq)? • Metals that can displace H2 from water
→ Fe2+/Fe is below and Cu2+/Cu /H+
is above H2 – The reduction of H2O to H2 is given by:
2H+ + 2e- → H2(g) (reduction) Eo = 0.00 V 2H2O(l) + 2e- → H2(g) + 2OH- E = -0.42 V
+
Fe(s) → Fe2+ + 2e- (oxidation) Eo = –0.44 V – The value of E is for pH = 7 (nonstandard state)
2H+ + 2e- + Fe(s) → H2(g) + Fe2+ + 2e- ⇒ Metals that are below H2O/H2,OH- in Appendix D
can displace H2 from water at standard conditions
Eocell = Eocath – Eoanod = 0.00 – (–0.44) = +0.44 V
⇒ Metals that have Eometal < -0.42 can displace H2
⇒ Eocell > 0 → spontaneous (Fe dissolves in HCl)
from water at pH = 7
2H+ + 2e- → H2(g) (reduction) Eo = 0.00 V Example: Potassium, K, dissolves readily in water
+
Cu(s) → Cu2+ + 2e- (oxidation) Eo = +0.34 V 2H2O(l) + 2e- → H2(g) + 2OH- (reduction) E = –0.42 V
2H + 2e + Cu(s) → H2(g) + Cu2+ + 2e-
+ -
K(s) → K+ + e- ×2 (oxidation) Eo = –2.93 V
Eocell = Eocath – Eoanod = 0.00 – (+0.34) = –0.34 V 2H2O(l) + 2e + 2K(s) → H2(g) + 2OH- + 2K+ + 2e-
-
⇒ Eocell < 0 → non-spontaneous (Cu doesn’t dissolve) Eocell = –0.42 – (–2.93) = +2.51 V > 0 (spontaneous)
21.4 Free Energy and Electrical Work – ∆Gr, ∆Gro, Ecell, and Eocell are all dependent on T
(superscripts, T, are omitted for simplicity)
Relationship Between Ecell and ∆Gr – ∆Gr, ∆Gro are extensive properties
• Electrical work (w) – Ecell, Eocell, E, and Eo are intensive properties
w = (charge transferred)×(voltage) ⇒ If a redox equation is multiplied by a number, ∆G
– n → # mol e- transferred (charge transferred) = nF is also multiplied, but E is not
– F → charge of 1 mol e- (voltage) = Ecell Example: Using Eo values from appendix D,
⇒ w = – nFEcell (w < 0 since the system does work) calculate ∆Gro at 298 K for the reaction:
– ∆G is the maximum work the system can do, so 2Cr3+ + 2Br- → 2Cr2+ + Br2(l)
∆G = wmax → Find the redox couples in Appendix D (298 K):
⇒ If the process is carried out reversibly (w = wmax) Cr3+ + e- → Cr2+ Eo = -0.41 V
∆Gr = – nFEcell and ∆Gro = – nFEocell Br2(l) + 2e → 2Br
- - Eo = +1.06 V
→ F = 96485 C/mol → Faraday constant → Invert the 2nd half-reaction to match the overall eq.
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Example: Using Eo values from appendix D, Interrelationship between ∆Gro, Eocell, and K
calculate K at 298 K for the reaction:
2Cr2+ + Br2(l) → 2Cr3+ + 2Br-
→ This is the reverse of the reaction in the previous
example (Eocell = -1.47 V from previous example)
→ Ereverse = - Eforward
⇒ Eocell = -(-1.47 V) = +1.47 V and n = 2
o
nE cell 2×1.47
⇒K =e 0.0257
=e 0.0257
= 4.8 × 1049
→Eocell > 0 → the reaction is spontaneous at standard
conditions
→ K >> 1 → the products are favored at equilibrium
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Concentration Cells 0.0257 0.0592 0.1
E cell = E cell
o
− logQ = 0 − log
• Concentration cell – contains the same redox n 2 1.0
couple in both the anode and cathode half-cells 0.0592
E = 0− (− 1) = 0 + 0.0296 = 0.0296 V
– The anode and cathode are the same 2
⇒ Eocell = Eocathode – Eoanode = 0
– The concentrations of the components are different in
the two half-cells ⇒ Ecell = 0 – (RT/nF) ln Q ≠ 0
Example: Cu2+/Cu concentration cell (EoCu= 0.34 V)
Cu2+(1.0 M) + 2e- → Cu(s) (cathode,reduction)
+
Cu(s) → Cu (0.1 M) + 2e
2+ - (anode,oxidation)
Cu2+(1.0 M) + 2e- + Cu(s) → Cu(s) + Cu2+(0.1 M) + 2e-
Eocell = EoCu – EoCu = +0.34 – (+0.34) = 0 V → The cell continues to work until [Cu2+] is equalized in the
two half-cells and Ecell decreases to zero
8
• Fuel cells – use combustion reactions 21.6 Corrosion
– The cell must be continuously provided with fuel and • Unwanted oxidation of metals in the environment
oxygen (flow cells) – If the metal (M) is in contact with water
Example: The hydrogen fuel cell Cathode, reduction:
2H2O(l) + 2e- → H2(g) + 2OH- Eo = -0.83 V
Eocell = Eocath - Eoanod (at pH = 7) → E = -0.42 V
Anode, oxidation:
Eocell = 1.23 – (0.00) ≈ 1.2 V
M(s) → Mn+ + ne- Eo < -0.42 V
PH 2O ⇒Any metal with Eo < -0.42 V can be oxidized by H2O
Q= Cathode, reduction:
PH 2 PO12/ 2
O2(g) + 4H+ + 4e- → 2H2O(l) Eo = +1.23 V
0.0257 (at pH = 7) → E = +0.82 V
E cell = E cell
o
− ln Q Anode, oxidation:
2 M(s) → Mn+ + ne- Eo < +0.82 V
⇒ ↑PH2 and ↑PO2 ⇒Any metal with Eo < +0.82 V can be oxidized by H2O
leads to ↑Ecell in the presence of O2
9
¾Electrolytic cells act in reverse (non-spontaneous) • Electrolysis – the passage of electrical current
direction compared to galvanic cells through an electrolyte by applying external
¾Eocell < 0 and ∆G > 0 (non-spontaneous reaction) voltage (the process in electrolytic cells)
¾The anode is positive and the cathode is negative
– Electrolysis causes a non-spontaneous reaction
¾There are some similarities between electrolytic
(often a splitting of a substance to its elements)
and galvanic cells
¾Oxidation is always on the anode and reduction is – The applied voltage must be greater than the cell
always on the cathode potential of the reverse spontaneous reaction
¾Electrons always flow from anode toward cathode – The electrolyte can be a molten salt or an aqueous
electrolyte solution
– Salt bridges are often not necessary
– During electrolysis the cations are attracted to the
cathode (negative) and the anions are attracted to
the anode (positive)
• Electrolysis of mixed molten salts →Al3+ is the stronger oxidizing agent because Al is
– The cation with higher Eo value (the stronger more EN than Na, so Al3+ gains electrons easier
oxidizing agent) is reduced at the cathode ⇒ Cathode half-reaction: Al3+(l) + 3e- → Al(l)
– The anion with lower Eo value (the stronger → Possible anode half-reactions (oxidation)
reducing agent) is oxidized at the anode 1) Oxidation of F- and 2) Oxidation of Cl-
Note: Eo values in appendix D are for aqueous ions →Cl- is the stronger reducing agent because Cl is
and can be used only as approximate guidance. less EN than F, so Cl- looses electrons easier
Instead, EN values can be used to estimate the ⇒ Anode half-reaction: 2Cl-(l) → Cl2(g) + 2e-
stronger oxidizing and reducing agents. Al3+(l) + 3e- → Al(l) (×2) cathode, reduction
Example: Predict the products of the electrolysis 2Cl-(l) → Cl2(g) + 2e- (×3) anode, oxidation
of a molten mixture of NaCl and AlF3 2Al3+(l) + 6e- + 6Cl-(l) → 2Al(l) + 2Cl2(g) + 6e-
→ Possible cathode half-reactions (reduction)
⇒ The products are Al(l) and Cl2(g)
1) Reduction of Na+ and 2) Reduction of Al3+
10
• Electrolysis of water → Ecell < 0 ⇒ non-spontaneous reaction
– Pure water is hard to electrolyze (low conductivity), ⇒ To drive the reaction, the external voltage must be
so a small amount of a non-reactive salt (NaNO3) is greater than 1.24 V
added (can be neglected)
• Overvoltage – the extra voltage (in the case of
→ H2O is reduced at the cathode: water over 1.24) needed to drive the reaction
2H2O(l) + 2e- → H2(g) + 2OH- Eo = -0.83 V – For H2O on most inert
(at pH = 7) → E = -0.42 V electrodes, the overvoltage is
→ H2O is oxidized at the anode: 0.4 to 0.6 V per electrode
2H2O(l) → O2(g) + 4H+ + 4e- Eo = +1.23 V → Cathode, reduction:
(at pH = 7) → E = +0.82 V E = -0.42 – 0.6 ≈ -1.0 V
4H2O(l) + 4e- + 2H2O(l) → 2H2(g) + 4OH- + O2(g) + 4H+ + 4e- → Anode, oxidation:
E = +0.82 + 0.6 ≈ +1.4 V
→ Overall: 2H2O(l) → 2H2(g) + O2(g)
→ Total:
Ecell = Ecath – Eanod = -0.42 – (+0.82 ) = -1.24 V Ecell = -1.0 – (+1.4 ) ≈ -2.4 V
11
¾ Electrolysis of aqueous solutions is often used The Stoichiometry of Electrolysis
for production or purification of less active • Faraday’s law – the amount of substance
elements produced on each electrode is directly
¾ Cations of less active metals (Cu, Ag, Au, Pt, …) proportional to the amount of charge transferred
are reduced on the cathode through the cell
¾ Anions of less active nonmetals (I2, Br2, …) are I → el. current t → time of electrolysis
oxidized on the anode (including Cl2 due to the Q → charge transferred n → # mol e- transferred
overvoltage of water) I = Q/t → Q = I×t I ×t
⇒ n=
¾ Cations of more active metals (Na, K, Mg, Ca, …) Q = nF → n = Q/F F
are not reduced (H2O is reduced to H2 instead) →
→ n is related to the amount of substance through
can’t be produced by electrolysis of aqueous soln.
the stoichiometry of the half-reaction
¾ Anions of more active nonmetals (F-) and
oxoanions of elements in their highest oxidation → Allows the determination of the amount of
state (NO3-, CO32-, SO42-, …) are not oxidized substance produced by measuring I and t
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