{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Chapter 6: Electrochemistry"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 1, Page no: 180"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import log10\n",
"\n",
"# Variables\n",
"T = 25 # C\n",
"E = 0.296 # V\n",
"Cu = 0.015 # M\n",
"\n",
"# Solution\n",
"Eo = E - 0.0296 * log10(Cu)\n",
"print \"The standard electrode potential is\", \"{:.3f}\".format(Eo), \"V\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The standard electrode potential is 0.350 V\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 2, Page no: 180"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import log10\n",
"\n",
"# Variables\n",
"T = 25 # C\n",
"Cu = 0.1 # M\n",
"Zn = 0.001 # M\n",
"Eo = 1.1\n",
"\n",
"# Solution\n",
"E = Eo + 0.0296 * log10(Cu / Zn)\n",
"print \"The emf of Daniell cell is\", \"{:.4f}\".format(E), \"V\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The emf of Daniell cell is 1.1592 V\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 3, Page no: 180"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import log10\n",
"\n",
"# Constant\n",
"R = 8.314 # J / K\n",
"F = 96500 # C / mol\n",
"\n",
"# Variables\n",
"Cu = 0.15 # M\n",
"Eo = 0.34 # V\n",
"T = 298 # K\n",
"n = 2 # moles\n",
"\n",
"# Solution\n",
"E = Eo + (2.303 * R * T) / (n * F) * log10(Cu)\n",
"print \"The single electrode potential for copper metal is\", \"{:.4f}\".format(E),\n",
"print \"V\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The single electrode potential for copper metal is 0.3156 V\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 4, Page no: 181"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variable\n",
"Eo_Cu = 0.3370 # Cu+2 -> Cu\n",
"Eo_Zn = - 0.7630 # Zn -> Zn +2\n",
"\n",
"# Solution\n",
"Eo_cell = Eo_Cu - Eo_Zn\n",
"print \"Daniel cell is, Zn | Zn +2 || Cu+2 | Cu\"\n",
"print \"Eo (cell) is\", Eo_cell, \"V\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Daniel cell is, Zn | Zn +2 || Cu+2 | Cu\n",
"Eo (cell) is 1.1 V\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 5, Page no: 181"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variable\n",
"Eo_Cu = 0.3370 # Cu+2 -> Cu\n",
"Eo_Cd = - 0.4003 # Cd -> Cd +2\n",
"\n",
"# Solution\n",
"Eo_cell = Eo_Cu - Eo_Cd\n",
"print \"Cell is, Cd | Cd +2 || Cu+2 | Cu\"\n",
"print \"Eo (cell) is\", Eo_cell, \"V\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Cell is, Cd | Cd +2 || Cu+2 | Cu\n",
"Eo (cell) is 0.7373 V\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 6, Page no: 181"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Constant\n",
"F = 96500 # C / mol\n",
"\n",
"# Variables\n",
"n = 2\n",
"T = 25 # C\n",
"Eo_Ag = 0.80 # Ag+ / Ag\n",
"Eo_Ni = - 0.24 # Ni+2 / Ni\n",
"\n",
"# Solution\n",
"Eo_Cell = Eo_Ag - Eo_Ni\n",
"print \"Standard free energy change,\"\n",
"delta_Go = - n * F * Eo_Cell\n",
"print delta_Go, \"J / mol\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Standard free energy change,\n",
"-200720.0 J / mol\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 7, Page no: 181"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Solution\n",
"print \"-------------------------\"\n",
"print \"Reduction half reaction:\",\n",
"print \"2Fe+3 + 2e- --> 2Fe+2\"\n",
"print \"Oxidation half reaction:\",\n",
"print \"Fe - 2e- --> Fe+2\"\n",
"print \"Overall cell reaction :\",\n",
"print \"2Fe+3 + Fe --> 3Fe+2\"\n",
"\n",
"print \"-------------------------\"\n",
"print \"Reduction half reaction:\",\n",
"print \"Hg+2 + 2e- --> Hg\"\n",
"print \"Oxidation half reaction:\",\n",
"print \"Zn - 2e- --> Zn+2\"\n",
"print \"Overall cell reaction :\",\n",
"print \"Hg+2 + Zn --> Hg + Zn+2\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"-------------------------\n",
"Reduction half reaction: 2Fe+3 + 2e- --> 2Fe+2\n",
"Oxidation half reaction: Fe - 2e- --> Fe+2\n",
"Overall cell reaction : 2Fe+3 + Fe --> 3Fe+2\n",
"-------------------------\n",
"Reduction half reaction: Hg+2 + 2e- --> Hg\n",
"Oxidation half reaction: Zn - 2e- --> Zn+2\n",
"Overall cell reaction : Hg+2 + Zn --> Hg + Zn+2\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 8, Page no: 181"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Constant\n",
"F = 96500 # C / mol\n",
"\n",
"# Variables\n",
"E1o = - 2.48 # V\n",
"E2o = 1.61 # V\n",
"\n",
"# Solution\n",
"delta_G1 = - 3 * F * (- 2.48)\n",
"delta_G2 = - 1 * F * 1.61\n",
"print \"delta_G3 = delta_G1 + delta_G2\"\n",
"print \"delta_G3 = - 4 * F * E3o\"\n",
"E3o = (delta_G1 + delta_G2) / (- 4 * F)\n",
"print \"The reduction potential for the half-cell Pt/Ce, Ce+4\",\n",
"print \"is\", \"{:.4f}\".format(E3o), \"V\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"delta_G3 = delta_G1 + delta_G2\n",
"delta_G3 = - 4 * F * E3o\n",
"The reduction potential for the half-cell Pt/Ce, Ce+4 is -1.4575 V\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 9, Page no: 182"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Solution\n",
"print \"Anodic half reaction :\",\n",
"print \" Cd --> Cd+2 + 2e-\"\n",
"print \"Cathodic half reaction :\",\n",
"print \"2H+ + 2e- --> H2\"\n",
"print \"-\" * 50\n",
"print \"Overall cell reaction :\",\n",
"print \"Cd + 2H+ <--> Cd+2 + H2\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Anodic half reaction : Cd --> Cd+2 + 2e-\n",
"Cathodic half reaction : 2H+ + 2e- --> H2\n",
"--------------------------------------------------\n",
"Overall cell reaction : Cd + 2H+ <--> Cd+2 + H2\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 10, Page no: 182"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import log10\n",
"\n",
"# Variables\n",
"T = 25 # C\n",
"Cu = 0.1 # M\n",
"Zn = 0.001 # M\n",
"Eo = 1.1 # V\n",
"\n",
"# Solution\n",
"print \"Zn(s) | Zn+2 (0.001M) || Cu+2(0.1M) | Cu(s)\"\n",
"Ecell = Eo + 0.0592 / 2 * log10(Cu / Zn)\n",
"print \"The emf of a Daniell cell is\", \"{:.4f}\".format(Ecell), \"V\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Zn(s) | Zn+2 (0.001M) || Cu+2(0.1M) | Cu(s)\n",
"The emf of a Daniell cell is 1.1592 V\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 11, Page no: 182"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import log10\n",
"\n",
"# Variables\n",
"pH = 7 # O2\n",
"Eo = 1.229 # V\n",
"pO2 = 0.20 # bar\n",
"\n",
"# Solution\n",
"print \"Nearnst's equation at 25C is,\"\n",
"print \"E = Eo - (0.0592 / 2) * log(1 / ([H+]^2 * [pO2]^ (1/2)))\"\n",
"E = Eo - (0.0592 / 2) * log10(1.0 / (((10 ** (- 7)) ** 2) * (pO2 ** (1 / 2.0))))\n",
"print \"The reduction potential for the reduction is\",\n",
"print \"{:.3f}\".format(E), \"V\"\n",
"# descrepency due to calculation error in the book"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Nearnst's equation at 25C is,\n",
"E = Eo - (0.0592 / 2) * log(1 / ([H+]^2 * [pO2]^ (1/2)))\n",
"The reduction potential for the reduction is 0.804 V\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 12, Page no: 183"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"E_KCl = 0.2415 # V\n",
"E_cell = 0.445 # V\n",
"\n",
"\n",
"# Solution\n",
"print \"Emf of the cell is\"\n",
"print \"At 25C,\"\n",
"print \"E = Er - El = Eref - ((RT)/ F) * ln H+\"\n",
"pH = (E_cell - E_KCl) / 0.059\n",
"\n",
"Eo_cell = - 0.8277 # V\n",
"print \"Thus, equilibrium constant for the reaction\"\n",
"print \"\\t 2H2O --> H3O+ + OH- may be calculated as\"\n",
"K = 10 ** (Eo_cell / 0.0591)\n",
"print \"K =\", \"{:.0e}\".format(K)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Emf of the cell is\n",
"At 25C,\n",
"E = Er - El = Eref - ((RT)/ F) * ln H+\n",
"Thus, equilibrium constant for the reaction\n",
"\t 2H2O --> H3O+ + OH- may be calculated as\n",
"K = 1e-14\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 13, Page no: 183"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"EoSn = 0.15 # V\n",
"EoCr = - 0.74 # V\n",
"\n",
"# Solution\n",
"print \"3Sn+4 + 2Cr --> 3Sn+2 + 2Cr+3\"\n",
"Eo_cell = EoSn - EoCr\n",
"n = 6\n",
"K = 10 ** (n * Eo_cell / 0.0591)\n",
"print \"The equillibrium constant for th reaction is\", \"{:.2e}\".format(K)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"3Sn+4 + 2Cr --> 3Sn+2 + 2Cr+3\n",
"The equillibrium constant for th reaction is 2.27e+90\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 14, Page no: 184"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"T = 25 # C\n",
"Eo = - 0.8277 # V\n",
"\n",
"# Solution\n",
"print \"The reversible reaction,\"\n",
"print \"2H2O <--> H3O+ + OH-\"\n",
"print \"May be divided into two parts.\"\n",
"print \"2H2O + e- --> 1/2 H2 + OH- (cathode) Eo = -0.8277 V\"\n",
"print \"H2O + 1/2 H2 --> H3O+ + e- (anode) Eo = 0\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The reversible reaction,\n",
"2H2O <--> H3O+ + OH-\n",
"May be divided into two parts.\n",
"2H2O + e- --> 1/2 H2 + OH- (cathode) Eo = -0.8277 V\n",
"H2O + 1/2 H2 --> H3O+ + e- (anode) Eo = 0\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 15, Page no: 184"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"E = 0.4 # V\n",
"\n",
"# Solution\n",
"print \"The cell is Pt(H2) | H+, pH2 = 1 atm\"\n",
"print \"The cell reaction is\"\n",
"print \"1/2 H2 --> H+ + e-\"\n",
"pH = E / 0.0591\n",
"print \"pH =\", \"{:.3f}\".format(pH)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The cell is Pt(H2) | H+, pH2 = 1 atm\n",
"The cell reaction is\n",
"1/2 H2 --> H+ + e-\n",
"pH = 6.768\n"
]
}
],
"prompt_number": 4
}
],
"metadata": {}
}
]
}