Chapter 21 Electrochemistry: Fundamentals Key Points About Redox Reactions 1. Oxidation (electron loss) always accompanies reduction (electron gain). 2. The oxidizing agent is reduced, and the reducing agent is oxidized. 3. The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent.
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PROBLEM: 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.
PLAN:
SOLUTION:
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).
Put the equations together in varying combinations so as to produce (+) E0
cell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of E0. Balance the number of electrons gained and lost without changing the E0.
In ranking the strengths, compare the combinations in terms of E0cell.
Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength
• Conventionally, the half cell potential refers to its reduction half-reaction.
• Using standard H2 reference electrode, other Eo half-cell can be measured and used for ranking
the oxidizing agent or reducing agent.
• Spontaneous redox reactions combine stronger oxidizing and reducing agent to form weaker ones.
• Spontaneous reaction is indicated negative ∆G and positive ∆E,
∆G = - nF∆E.
• We can determine K using ∆E, ∆Go = -nF∆Eo = - RTlnK.
Relative Reactivities (Activities) of Metals
1. Metals that can displace H from acid
2. Metals that cannot displace H from acid
3. Metals that can displace H from water
4. Metals that can displace other metals from solution
can displace Hfrom water
LiKBaCaNa
stre
ngt
h as
red
ucin
g ag
ents
can displace Hfrom steam
MgAlMnZnCrFeCd
H2
cannot displace H from any source
CuHg AgAu
can displace Hfrom acid
CoNiSnPb
G0
E0cell K
G0 K
Reaction at standard-state
conditionsE0cell
The interrelationship of G0, E0, and K.
< 0 spontaneous
at equilibrium
nonspontaneous
0
> 0
> 0
0
< 0
> 1
1
< 1
G0 = -RT lnKG0 = -nFEo
cell
E0cell = -RT lnK
nF
By substituting standard state values into E0
cell, we get
E0cell = (0.0592V/n) log K (at 298 K)
The Effect of Concentration on Cell Potential
G = G0 + RT ln Q
-nF Ecell = -nF Ecell + RT ln Q
Ecell = E0cell - ln Q
RT
nF
•When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell
•When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell
•When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell
Ecell = E0cell -
log Q0.0592
n
Nernst equation
Calculating K and G0 from E0cell
PLAN:
SOLUTION:
PROBLEM: Lead can displace silver from solution:
As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and G0 at 298 K for this reaction.
Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s)
Break the reaction into half-reactions, find the E0 for each half-reaction and then the E0
cell. Substitute into the equations found on slide
E0 = -0.13V Anode
E0 = 0.80V Cathode
E0cell = E0
cathode – E0anode = 0.93V
Ag+(aq) + e- Ag(s)
Pb2+(aq) + 2e- Pb(s)
Ag+(aq) + e- Ag(s)
E0cell = log K
0.592V
n
log K =
K = 2.6x1031
n x E0cell
0.592V(2)(0.93V)
0.592V=
G0 = -nFE0cell
= -(2)(96.5kJ/mol*V)(0.93V)
G0 = -1.8x102kJ
2X
E0cell = - (RT/n F) ln K
Using the Nernst Equation to Calculate Ecell
PROBLEM: In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions:
PLAN:
SOLUTION:
[Zn2+] = 0.010M [H+] = 2.5M P = 0.30atmH2
Calculate Ecell at 298 K.
Find E0cell and Q in order to use the Nernst equation.
Determining E0cell :
E0 = 0.00V2H+(aq) + 2e- H2(g)
E0 = -0.76VZn2+(aq) + 2e- Zn(s)
Zn(s) Zn2+(aq) + 2e- E0 = +0.76V
Q = P x [Zn2+]
H2
[H+]2
Q = 4.8x10-4
Q = (0.30)(0.010)
(2.5)2
Ecell = E0cell -
0.0592V
nlog Q
Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V
Diagramming Voltaic Cells
PROBLEM: Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Zn bar in a Zn(NO3)2 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Zn electrode is negative relative to the Ag electrode.
PLAN:
SOLUTION:
Identify the redox reactions
Write each half-reaction.
Associate the (-)(Zn) pole with the anode (oxidation) and the (+) (Ag) pole with the cathode (reduction).
• When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell forward reaction
• When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell Reverse reaction
• When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell Equilibrium
Ecell = E0cell -
log Q0.0592
n
Nernst equation
Cell operates with all components at standard states. Most cells are starting at Non-standard state.
Sample Problem Using the Nernst Equation to Calculate Ecell
PROBLEM: In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions:
PLAN:
SOLUTION:
[Zn2+] = 0.010M [H+] = 2.5M P = 0.30atmH2
Calculate Ecell at 298 K.
Find E0cell and Q in order to use the Nernst equation.
Determining E0cell :
E0 = 0.00V cathode2H+(aq) + 2e- H2(g)
E0 = -0.76V anodeZn2+(aq) + 2e- Zn(s)Q =
P x [Zn2+]H2
[H+]2
Q = 4.8x10-4
Q = (0.30)(0.010)
(2.5)2
Ecell = E0cell -
0.0592
nlog Q
Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V
Ecell = E0cell - ln Q
RT
nF
Ecell0 = E0
c-E0a = 0.00-(-0.76)V = 0.76 V
2H+(aq) + Zn (s) H2(g) + Zn2+ (aq)
Ecell = E0cell -
log Q0.0592
n
Summary
• A voltaic cell contains of oxidation (anode) and reduction (cathode) half-cells, connected by a salt bridge.
• The salt bridge provides ions to maintain the charge balance when the cell operates.
• Electrons move from anode to cathode while cation moves from salt bridge to the cathode half cell.
• The output of a cell is called cell potential (Ecell) and is measured in volts.
• When all substances are in standard states, the cell potential is the standard cell potential (Eo