School of Chemistry, University of KwaZulu-Natal, Westville Campus, Durban
CHEM 120R
Electrochemistry Tutorial
1. Consider the following reaction:
ClO3- + H+  ClO4- + ClO2 + Cl2 + O2
(a)
Assign oxidation numbers to all the atoms or ions in the equation
Cl(+5)O(-2) + H+(+1)  Cl(+7)O(-2)- + Cl(+4)O(-2) + Cl2(0) + O2(0)
(b)
(i)
Which substance is oxidized? ClO3(both Cl and O increase in ON)
(ii)
Which substance is reduced? ClO3( Cl decreases in ON)
(iii)
Which substance is the oxidizing agent? ClO3-
(iv)
Which substance is the reducing agent? ClO3-
(v)
This type of redox reaction is also called a disproportionation
reaction.
Same substance (in this case ClO3-) is oxidized and reduced.
2.
For each of the following unbalanced equations, write the half-reactions for
oxidation and for reduction and balance the overall cell reactions.
(i)
Hg2+ + Cu  Hg + Cu2+
oxidation ½- reaction:
Cu  Cu2+ + 2e-
reduction ½- reaction: Hg2+ + 2e-  Hg
overall:
(ii)
Hg2+ + Cu  Hg + Cu2+
MnO2 + Cl-  Mn2+ + Cl2
oxidation ½- reaction:
2Cl-  Cl2 + 2e-
reduction ½- reaction: MnO2 + 2e- + 4H+  Mn2+ + 2H2O
overall:
MnO2 + 2Cl- + 4H+  Mn2+ + Cl2 + 2H2O
1
(iii)
3.
Sn2+ + O2  Sn4+ + H2O
oxidation ½- reaction:
Sn2+  Sn4+ + 2e-
reduction ½- reaction:
O2 + 4H+ + 4e-  2H2O
overall:
2Sn2+ + O2 + 4H+  2Sn4+ + 2H2O
x2
What is the potential of the cell involving this reaction?
Zn + 2Ag+  Zn2++ 2Ag
(Assume all substances in their standard states.)
Standard Reduction Potentials
Zn2+ + 2e–  Zn E0 = –0.76 V
Ag+ + e–  Ag E0 = +0.80 V
Eocell = Eored - Eoox = +0.80 V – (-0.76) V = 1.56 V
4. For each of the following cells :
(a)
Ag  Ag+  Cd2+  Cd and
(i)
Write down the net reaction in the direction consistent with the way the
(b)
Zn  Zn2+  Co2+  Co
cell is written.
a) 2Ag + Cd2+ → Cd + 2Ag+
b)Zn + Co2+ → Co + Zn2+
(ii)
Write the half-reactions for the anode and cathode processes.
anode:
a)
Ag → Ag+ + e-
b)
Zn → Zn2+ + 2e-
cathode:
a)
Cd2+ + 2e- → Cd
b)
Co2+ + 2e- → Co
2
iii)
Find the standard cell potential at 25oC.
(a) Eocell = Eored - Eoox = -0.40 V – 0.80 V = -1.20 V
(b) Eocell = Eored - Eoox = -0.28 V – (- 0.76 V) = 0.48 V
iv)
State whether the standard cell reaction actually occurs as given or in the
reverse direction.
5.
(a)
reverse
(b)
foward
Consult a table of standard reduction potentials and determine which of the
following reactions are spontaneous under standard electrochemical conditions :
6.
(i)
Mn(s) + 2H+(aq)  H2(g) + Mn2+(aq)
(ii)
Cl2(g) + 2Br-(aq) 
Will
(i)
Br2(l) + 2Cl-(aq)
Fe3+ oxidize Sn2+ to Sn4+ in acidic solution?
Assume that Fe3+ oxidize Sn2+ to Sn4+ in acidic solution
oxidation ½- reaction: Sn2+ → Sn4+ + 2ereduction ½- reaction: Fe3+ + e- → Fe2+
Eocell = Eored - Eoox = +0.77 V – 0.15 V = +0.62 V
Therefore, Fe3+ will oxidize Sn2+ to Sn4+ in acidic solution
(ii)
permanganate ions oxidize arsenous acid, H3AsO3 to arsenious acid,
H3AsO4 in acidic solution ?
(iii)
dichromate ions oxidize Mn2+ to MnO4- in acidic solution ?
7. Calculate the potential associated with the following half-reaction when the
concentration of the cobalt(II) ion is 1.0 x 10-3 M.
Co(s)  Co2+ + 2eEcell = -0.28 V - {(0.0592 V/2)log(1 × 10-3)} = -0.28 V + 0.0888 V = -0.191 V
3
8.
Consider the cell represented by the notation :
Zn(s)  ZnCl2(aq)  Cl2(g,1atm); Cl-(aq)  C(s)
Calculate the Eo for the cell and calculate the E for the cell when the concentration
of the ZnCl2 is 0.15 M.
overall cell reaction: Zn + Cl2 → 2Cl- + Zn2+
Eocell = Eored - Eoox = 1.36 V – (-0.76 V) = 2.12 V
Ecell = 2.12 V - {(0.0592 V/2)log(0.15)(0.30)2} = 2.12 V + 0.055 V = 2.175 V
9. Consider the electrochemical cell represented by:
Zn(s)  Zn2+  Fe3+  Fe(s)
(i)
Write the ion-electron equations for the half-reactions and the overall cell
reaction.
Zn  Zn2+ + 2eFe3+ + 3e-  Fe
Zn + Fe3+  Zn2+ + Fe
(ii)
Determine the standard potential for the reaction.
Eo = -0.04 V - (-0.76 V) = 0.72 V
Determine E for the cell when the concentration of Fe3+ is 10.0 M and that of
(iii)
Zn2+ is 1.00 x 10-3 M.
Ecell = 0.72 V - {(0.0592 V/6)log(1 × 10-3)3/(10)2}
= 0.72 V + 0.109 V = 0.829 V
10.
Given the following half-cell reactions with their standard potentials:
(i)
Ag+(aq) + e-  Ag(s)
:
Eo = +0.80 V
Cd2+(aq) + 2e-  Cd(s)
:
Eo = -0.40 V
Write the cell notation, according to convention, for the galvanic cell
formed by coupling the above half-cells.
Cd Cd2+  Ag+  Ag
(ii)
Write the electrode reactions for this electrochemical cell
Cd  Cd2+ + 2e- and Ag+ + e-  Ag
(iii)
Write the overall cell reaction giving a balanced equation.
4
Cd + 2Ag+  Ag + Cd2+
(iv)
Calculate the e.m.f (Eocell.) of the cell.
Eo = 0.80 V – (-0.40 V) = 1.20 V
(v)
Calculate the free energy change for the overall cell reaction.
ΔGo = - nFEo = - 2(96485 C mol-1)(1.20 V) = - 2.32 x 105 J mol-1
(vi)
Calculate the e.m.f. of the cell (Ecell) at 25oC when [Ag+] = 0.100 M and
[Cd2+] = 0.010 M
Ecell = 1.20 V - {(0.0592 V/2)log(0.010)/(0.1)2} = 1.20 V
(vii)
Calculate the equilibrium constant, K, for the reaction at 25oC.
ΔGo = - RTlnK,
lnK = (2.32 × 105 J mol-1)/(8.315 J mol-1 K-1)(298 K) = 93.63
K = 4.6 x 1040
11. For the following cell :
Zn  Zn2+  Co2+  Co
(iii)
Write down the net reaction in the direction consistent with the way the
cell is written.
Zn + Co2+ → Co + Zn2+
(iv)
Write the half-reactions for the anode and cathode processes.
anode:
Zn → Zn2+ + 2ecathode:
Co2+ + 2e- → Co
5
(v)
Find the standard cell potential at 25oC.
Eocell = Eored - Eoox = -0.28 V – (- 0.76 V) = 0.48 V
(vi)
State whether the standard cell reaction actually occurs as given or in the
reverse direction.
foward
12.
Will Fe3+ oxidize Sn2+ to Sn4+ in acidic solution?
Assume that Fe3+ oxidize Sn2+ to Sn4+ in acidic solution
oxidation ½- reaction: Sn2+ → Sn4+ + 2ereduction ½- reaction: Fe3+ + e- → Fe2+
Eocell = Eored - Eoox = +0.77 V – 0.15 V = +0.62 V
Therefore, Fe3+ will oxidize Sn2+ to Sn4+ in acidic solution
13.
Calculate the potential associated with the following half-reaction when the
concentration of the cobalt(II) ion is 1.5 x 10-2 M.
Co(s)  Co2+ + 2eEcell = -0.28 V - {(0.0592 V/2)log(1.5 × 10-2)}
= -0.28 V – (-0.0540 V) = -0.226 V
14.
Consider the cell represented by the notation :
Zn(s)  ZnCl2(aq)  Cl2(g,1atm); Cl-(aq)  C(s)
Calculate the Eo for the cell and calculate the E for the cell when the concentration
of the ZnCl2 is 0.10 M.
overall cell reaction: Zn + Cl2 → 2Cl- + Zn2+
Eocell = Eored - Eoox = 1.36 V – (-0.76 V) = 2.12 V
Ecell = 2.12 V - {(0.0592 V/2)log(0.10)(0.20)2} = 2.12 V - (-0.0710 V)
= 2.191 V
6
15. Predict whether the following reaction will occur as written at 298 K, assuming
[Co2+] = 0.15 M and [Fe2+] = 0.68 M. Write the anode and cathode half reactions.
Co(s) + Fe2+ (aq)  Co2+(aq) + Fe(s)
EoFe2+ /Fe = -0.44 V
EoCo2+/Co = -0.28 V
anode:
Co(s) → Co2+ + 2ecathode:
Fe2+ + 2e- → Fe(s)
The standard cell potential
Eocell = Eoreduction - Eooxidation = -0.44 V – (-0.28 V) = -0.16 V
The reactant quotient, Q for the reaction is:
[Co2+] / [Fe2+] = (0.15/0.68) = 0.22.
E = Eo - 0.0592/n log Q = - 0.16 V – (0.0592/2)log 0.22 = -0.14 V
The negative E value indicates that the reaction is not spontaneous as
written under the conditions described.
16.
Consider the following reaction at 25oC.
Sn(s) + 2Cu2+ (aq)  Sn2+(aq) + 2Cu+(aq)
Eo (Sn2+/Sn) = -0.14 V
Eo (Cu2+/Cu+) = 0.15 V
(i) Write the half-reactions for the anode and cathode processes.
anode:
Sn(s) → Sn2+ + 2ecathode:
2Cu2+ + 2e- → 2Cu+(aq)
7
(ii) Find the standard cell potential at 25 oC.
Eocell = Eoreduction - Eooxidation = 0.15 V – (-0.14 V) = 0.29 V
(iii) Calculate the equilibrium constant, K, for the reaction at 25 oC.
ΔGo = - RTlnK, (determine ΔGo)
lnK = (ΔGo)/(8.315 J mol-1 K-1)(298 K)
K = 6 x 109
17.
Calculate the standard free energy change for the following reaction at 25 ˚C.
2Au(s) + Ca2+ (aq,1M)  2Au3+(aq,1M) + Ca(s)
EoAu3+ /Au = 1.50 V
EoCa2+/Ca = -2.87 V
i) Write the half-reactions for the anode and cathode processes.
anode:
2Au(s) → 2Au3+ + 6ecathode:
3Ca2+ + 6e- → 3Ca
Find the standard cell potential at 25oC.
ii)
Eocell = Eoreduction - Eooxidation = -2.87 V – (1.50 V) = -4.37 V
iii)
Calculate the free energy change for the overall cell reaction.
o
ΔG = - nFEo = - 6(96485 J/V mol-1)(-4.37 V) = 2.53 × 106 J mol-1
= 2.53 × 103 kJ mol-1
Additional Problems
1.
Which is the direction of spontaneous change for the following reaction – to the
left or to the right?
8
2 Cr
3
 6 Fe
3
?
2
2
 7 H2O 

Cr2O7  14 H  6 Fe
Answer
The two ½-reactions of interest are
Fe
3
 e  Fe
2
E0 = +0.77 V

2-
Cr2O7  14 H  6 e  2 Cr
3
 7 H2O
E0 = +1.33 V
For the overall reaction equation to proceed to the right as written, the Fe3+ will be
reduced and the Cr3+ will be oxidised thus
0
0
3
2
0
3
2-
ΔE  E ( Fe /Fe ) - E (Cr2O7 /Cr )
  0.77 V - 1.33 V
 - 0.56 V
Therefore this reaction will not go spontaneously.
If the overall reaction proceeds to the left then
0
0
2-
ΔE  E (Cr2O7 /Cr
3
0
3
2
) - E ( Fe /Fe )
 1.33 - 0.77 V
  0.56 V
This means that the spontaneous reaction will be
2-

Cr2O7  14 H  6 Fe
2
 2 Cr
3
 6 Fe
3
 7 H2O
i.e the dichromate ion will oxidise Fe2+ to Fe3+ and be itself reduced to Cr3+.
2.
Calculate the equilibrium constant for the reaction between MnO4 and Fe2+ in
acidic aqueous solution at 25 oC.
The reaction equation is

2+
MnO4 + 5Fe
+ 8H+ ⇌ Mn2+ + 5Fe3+ + 4H2O
9
Answer
The number of electrons being transferred, n , equals 5.
If we look up the table of relevant Eo values we will find
Eo ( MnO4 /Mn2+) = 1.51 V
Eo (Fe3+/Fe2+) = 0.77 V
o
The manganese couple is
o
E red and the iron couple is E ox . Therefore
o
o
Eo = E red - E ox
= 1.51 – 0.77
= 0.74 V
Now we have all the information we need to calculate Kc.
ln Kc = nFE/RT
5 x 96485 C mol
=
8.315 J mol
1
K
1
 0.74 V
1
 298 K
Let’s make sure we know what’s happening with the units –
A coloumb, C, the measurement of charge is
C = A
x s
Amp x second
(from the equation: charge = current x time).
A volt, V, the measurement of Eo is
V = J s-1 A-1
from the equation: potential difference = energy/charge)
 lnK C 
5 x 96485 A s mol
8.315 J mol
1
1
 0.74 J s
K
1
1
 298 K
= 144
and KC = 3.7 × 1062
10
A
1