Chapter 6: Electrochemistry

Problem: 1, Page no: 180

In [2]:
from math import log10

# Variables
T = 25              # C
E = 0.296           # V
Cu = 0.015          # M

# Solution
Eo = E - 0.0296 * log10(Cu)
print "The standard electrode potential is", "{:.3f}".format(Eo), "V"

The standard electrode potential is 0.350 V

Problem: 2, Page no: 180

In [4]:
from math import log10

# Variables
T = 25                  # C
Cu = 0.1                # M
Zn = 0.001              # M
Eo = 1.1

# Solution
E = Eo + 0.0296 * log10(Cu / Zn)
print "The emf of Daniell cell is", "{:.4f}".format(E), "V"

The emf of Daniell cell is 1.1592 V

Problem: 3, Page no: 180

In [1]:
from math import log10

# Constant
R = 8.314               # J / K
F = 96500               # C / mol

# Variables
Cu = 0.15               # M
Eo = 0.34               # V
T = 298                 # K
n = 2                   # moles

# Solution
E = Eo + (2.303 * R * T) / (n * F) * log10(Cu)
print "The single electrode potential for copper metal is", "{:.4f}".format(E),
print "V"

The single electrode potential for copper metal is 0.3156 V

Problem: 4, Page no: 181

In [2]:
# Variable
Eo_Cu = 0.3370          # Cu+2 -> Cu
Eo_Zn = - 0.7630        # Zn -> Zn +2

# Solution
Eo_cell = Eo_Cu - Eo_Zn
print "Daniel cell is, Zn | Zn +2 || Cu+2 | Cu"
print "Eo (cell) is", Eo_cell, "V"

Daniel cell is, Zn | Zn +2 || Cu+2 | Cu
Eo (cell) is 1.1 V

Problem: 5, Page no: 181

In [4]:
# Variable
Eo_Cu = 0.3370          # Cu+2 -> Cu
Eo_Cd = - 0.4003        # Cd -> Cd +2

# Solution
Eo_cell = Eo_Cu - Eo_Cd
print "Cell is, Cd | Cd +2 || Cu+2 | Cu"
print "Eo (cell) is", Eo_cell, "V"

Cell is, Cd | Cd +2 || Cu+2 | Cu
Eo (cell) is 0.7373 V

Problem: 6, Page no: 181

In [5]:
# Constant
F = 96500               # C / mol

# Variables
n = 2
T = 25                  # C
Eo_Ag = 0.80            # Ag+ / Ag
Eo_Ni = - 0.24          # Ni+2 / Ni

# Solution
Eo_Cell = Eo_Ag - Eo_Ni
print "Standard free energy change,"
delta_Go = - n * F * Eo_Cell
print delta_Go, "J / mol"

Standard free energy change,
-200720.0 J / mol

Problem: 7, Page no: 181

In [6]:
# Solution
print "-------------------------"
print "Reduction half reaction:",
print "2Fe+3 + 2e- --> 2Fe+2"
print "Oxidation half reaction:",
print "Fe    - 2e- --> Fe+2"
print "Overall cell reaction  :",
print "2Fe+3 + Fe  --> 3Fe+2"

print "-------------------------"
print "Reduction half reaction:",
print "Hg+2 + 2e- --> Hg"
print "Oxidation half reaction:",
print "Zn   - 2e- --> Zn+2"
print "Overall cell reaction  :",
print "Hg+2 + Zn  --> Hg + Zn+2"

-------------------------
Reduction half reaction: 2Fe+3 + 2e- --> 2Fe+2
Oxidation half reaction: Fe    - 2e- --> Fe+2
Overall cell reaction  : 2Fe+3 + Fe  --> 3Fe+2
-------------------------
Reduction half reaction: Hg+2 + 2e- --> Hg
Oxidation half reaction: Zn   - 2e- --> Zn+2
Overall cell reaction  : Hg+2 + Zn  --> Hg + Zn+2

Problem: 8, Page no: 181

In [7]:
# Constant
F = 96500               # C / mol

# Variables
E1o = - 2.48            # V
E2o = 1.61              # V

# Solution
delta_G1 = - 3 * F * (- 2.48)
delta_G2 = - 1 * F * 1.61
print "delta_G3 = delta_G1 + delta_G2"
print "delta_G3 = - 4 * F * E3o"
E3o = (delta_G1 + delta_G2) / (- 4 * F)
print "The reduction potential for the half-cell Pt/Ce, Ce+4",
print "is", "{:.4f}".format(E3o), "V"

delta_G3 = delta_G1 + delta_G2
delta_G3 = - 4 * F * E3o
The reduction potential for the half-cell Pt/Ce, Ce+4 is -1.4575 V

Problem: 9, Page no: 182

In [4]:
# Solution
print "Anodic half reaction   :",
print "      Cd   --> Cd+2 + 2e-"
print "Cathodic half reaction :",
print "2H+ + 2e-  --> H2"
print "-" * 50
print "Overall cell reaction  :",
print "Cd  + 2H+ <--> Cd+2 + H2"

Anodic half reaction   :       Cd   --> Cd+2 + 2e-
Cathodic half reaction : 2H+ + 2e-  --> H2
--------------------------------------------------
Overall cell reaction  : Cd  + 2H+ <--> Cd+2 + H2

Problem: 10, Page no: 182

In [5]:
from math import log10

# Variables
T = 25                  # C
Cu = 0.1                # M
Zn = 0.001              # M
Eo = 1.1                # V

# Solution
print "Zn(s) | Zn+2 (0.001M) || Cu+2(0.1M) | Cu(s)"
Ecell = Eo + 0.0592 / 2 * log10(Cu / Zn)
print "The emf of a Daniell cell is", "{:.4f}".format(Ecell), "V"

Zn(s) | Zn+2 (0.001M) || Cu+2(0.1M) | Cu(s)
The emf of a Daniell cell is 1.1592 V

Problem: 11, Page no: 182

In [6]:
from math import log10

# Variables
pH = 7                  # O2
Eo = 1.229              # V
pO2 = 0.20              # bar

# Solution
print "Nearnst's equation at 25C is,"
print "E = Eo - (0.0592 / 2) * log(1 / ([H+]^2 * [pO2]^ (1/2)))"
E = Eo - (0.0592 / 2) * log10(1.0 / (((10 ** (- 7)) ** 2) * (pO2 ** (1 / 2.0))))
print "The reduction potential for the reduction is",
print "{:.3f}".format(E), "V"
# descrepency due to calculation error in the book

Nearnst's equation at 25C is,
E = Eo - (0.0592 / 2) * log(1 / ([H+]^2 * [pO2]^ (1/2)))
The reduction potential for the reduction is 0.804 V

Problem: 12, Page no: 183

In [3]:
# Variables
E_KCl = 0.2415              # V
E_cell = 0.445              # V


# Solution
print "Emf of the cell is"
print "At 25C,"
print "E = Er - El = Eref - ((RT)/ F) * ln H+"
pH = (E_cell - E_KCl) / 0.059

Eo_cell = - 0.8277          # V
print "Thus, equilibrium constant for the reaction"
print "\t 2H2O --> H3O+  + OH- may be calculated as"
K = 10 ** (Eo_cell / 0.0591)
print "K =", "{:.0e}".format(K)

Emf of the cell is
At 25C,
E = Er - El = Eref - ((RT)/ F) * ln H+
Thus, equilibrium constant for the reaction
	 2H2O --> H3O+  + OH- may be calculated as
K = 1e-14

Problem: 13, Page no: 183

In [1]:
# Variables
EoSn = 0.15             # V
EoCr = - 0.74           # V

# Solution
print "3Sn+4  +  2Cr  -->  3Sn+2 + 2Cr+3"
Eo_cell = EoSn - EoCr
n = 6
K = 10 ** (n * Eo_cell / 0.0591)
print "The equillibrium constant for th reaction is", "{:.2e}".format(K)

3Sn+4  +  2Cr  -->  3Sn+2 + 2Cr+3
The equillibrium constant for th reaction is 2.27e+90

Problem: 14, Page no: 184

In [2]:
# Variables
T = 25                  # C
Eo = - 0.8277           # V

# Solution
print "The reversible reaction,"
print "2H2O  <--> H3O+ + OH-"
print "May be divided into two parts."
print "2H2O + e- --> 1/2 H2 + OH-    (cathode) Eo = -0.8277 V"
print "H2O  + 1/2 H2 --> H3O+ + e-   (anode) Eo = 0"

The reversible reaction,
2H2O  <--> H3O+ + OH-
May be divided into two parts.
2H2O + e- --> 1/2 H2 + OH-    (cathode) Eo = -0.8277 V
H2O  + 1/2 H2 --> H3O+ + e-   (anode) Eo = 0

Problem: 15, Page no: 184

In [4]:
# Variables
E = 0.4                 # V

# Solution
print "The cell is Pt(H2) | H+, pH2 = 1 atm"
print "The cell reaction is"
print "1/2 H2 --> H+ + e-"
pH = E / 0.0591
print "pH =", "{:.3f}".format(pH)

The cell is Pt(H2) | H+, pH2 = 1 atm
The cell reaction is
1/2 H2 --> H+ + e-
pH = 6.768