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+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9 - Electrochemistry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the potential of the cell\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "Gr=-math.pow(10,5) #kJ/mol\n",
+ "v=1\n",
+ "F=9.6485*10000. #C/mol\n",
+ "#calculations\n",
+ "E=-Gr/(v*F)\n",
+ "#results\n",
+ "print '%s %d %s' %(\"potential of the cell =\",E,\"V\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "potential of the cell = 1 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the equilibrium constant of the reaction\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "V=1.1 #V\n",
+ "F=9.6485*10000. #C/mol\n",
+ "R=8.314 #J/K mol\n",
+ "T=298.15 #K\n",
+ "#calculations\n",
+ "lnK=2*F*V/(R*T)\n",
+ "k=math.pow(math.e,(lnK))\n",
+ "#results\n",
+ "print '%s %.1e' %(\"Equilibrium constant =\",k)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equilibrium constant = 1.5e+37\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the acidity constant of the acid\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "lw=34.96 #mS m^2 /mol\n",
+ "la=4.09 #mS m^2 /mol\n",
+ "C=0.010 #M\n",
+ "K=1.65 #mS m^2 /mol\n",
+ "#calculations\n",
+ "lmd=lw+la\n",
+ "alpha=K/lmd\n",
+ "Ka=C*alpha*alpha\n",
+ "pKa=-math.log10(Ka)\n",
+ "#results\n",
+ "print '%s %.2f' %(\"Acidity constant of the acid = \",pKa)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Acidity constant of the acid = 4.75\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 - Pg 203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate if the reaction is favoring the products or not. Also calculate the E value\n",
+ "#Initialization of variables\n",
+ "ER=1.23 #V\n",
+ "EL=-0.44 #V\n",
+ "#calculations\n",
+ "E=ER-EL\n",
+ "#results\n",
+ "if(E>0):\n",
+ " print '%s %.2f %s' %(\"The reaction is favouring products and E is\",E,\"V\")\n",
+ "else:\n",
+ " print '%s %.2f %s' %(\"The reaction is not favouring products and E is\",E,\" V\")\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reaction is favouring products and E is 1.67 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 - Pg 203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the equilibrium constant \n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "ER=0.52 #V\n",
+ "EL=0.15 #V\n",
+ "#calculations\n",
+ "E=ER-EL\n",
+ "lnK=E*1000./(25.69)\n",
+ "K=math.exp(lnK)\n",
+ "#results\n",
+ "print '%s %.1e' %(\"Equilbrum constant K= \",K)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equilbrum constant K= 1.8e+06\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8 - Pg 205"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the biological standard potential\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "E0=-0.11 #V\n",
+ "H=math.pow(10,-7)\n",
+ "#calculations\n",
+ "pH=-math.log10(H)\n",
+ "E=E0-29.59*pH*math.pow(10,-3)\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"Biological standard potential =\",E,\"V\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Biological standard potential = -0.32 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9 - Pg 206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the equilibrium constant for the reaction\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "ER=-0.21 #V\n",
+ "EL=-0.6 #V\n",
+ "#calculations\n",
+ "E=ER-EL\n",
+ "lnK=2*E*1000./(25.69)\n",
+ "K=math.exp(lnK)\n",
+ "#results\n",
+ "print '%s %.1e' %(\"Equilibrium constant for the reaction = \",K)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equilibrium constant for the reaction = 1.5e+13\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 209"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the electric potential\n",
+ "#Initialization of variables\n",
+ "E1=2*(-0.340)\n",
+ "E2=-0.522 \n",
+ "#calculations\n",
+ "FE=-E1+E2\n",
+ "#results\n",
+ "print '%s %.3f %s' %(\"Electric potential =\",FE,\"V\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric potential = 0.158 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 - Pg 210"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Gibbs enthalpy, Standard entropy and enthalpy of the process\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "v=2\n",
+ "F=9.6485*10000. #C/mol\n",
+ "E=0.2684 #V\n",
+ "V1=0.2699 #V\n",
+ "V2=0.2669 #V\n",
+ "T1=293. #K\n",
+ "T=298. #K\n",
+ "T2=303. #K\n",
+ "#calculations\n",
+ "Gr= -v*F*E/1000.\n",
+ "Sr=v*F*(V2-V1)/(T2-T1)\n",
+ "Hr=Gr+T*Sr/1000.\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"Gibbs enthalpy =\",Gr,\"kJ/mol\")\n",
+ "print '%s %.1f %s' %(\"\\n Standard Entropy =\",Sr,\"J /K mol\")\n",
+ "print '%s %.1f %s' %(\"\\n Enthalpy =\",Hr,\"kJ/mol\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gibbs enthalpy = -51.79 kJ/mol\n",
+ "\n",
+ " Standard Entropy = -57.9 J /K mol\n",
+ "\n",
+ " Enthalpy = -69.0 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
generated by cgit (git 2.34.1) at 2024-12-20 07:27:34 +0000