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author | hardythe1 | 2015-04-07 16:03:32 +0530 |
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committer | hardythe1 | 2015-04-07 16:03:32 +0530 |
commit | ba4d7bfd00d268a9a5fbdee8938f9a4b5a04851d (patch) | |
tree | fcc60b142916d4a738807d3de7ed170b3f9818f6 /Chemistry/Chapter_19.ipynb | |
parent | 92cca121f959c6616e3da431c1e2d23c4fa5e886 (diff) | |
parent | 121f6027864eec574ad581705eb277e19120720c (diff) | |
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1 files changed, 0 insertions, 267 deletions
diff --git a/Chemistry/Chapter_19.ipynb b/Chemistry/Chapter_19.ipynb deleted file mode 100755 index c904a842..00000000 --- a/Chemistry/Chapter_19.ipynb +++ /dev/null @@ -1,267 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19:Electrochemistry"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:19.3,Page no:848"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "E0cathode=0.8 #standard electrode potential of cathode(Ag+/Ag), V\n",
- "E0anode=-2.37 #standard electrode potential of anode(Mg2+/Mg), V\n",
- "\n",
- "#Calculation\n",
- "E0cell=E0cathode-E0anode #standard emf of the cell, V\n",
- "\n",
- "#Result\n",
- "print\"The standard emf of the cell is :\",E0cell,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The standard emf of the cell is : 3.17 V\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:19.4,Page no:850"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "n=2.0 \n",
- "E0cathode=0.15 #standard electrode potential of cathode(Cu2+/Cu+), V\n",
- "E0anode=-0.14 #standard electrode potential of anode(Sn2+/Sn), V\n",
- "\n",
- "#Calculation\n",
- "E0cell=E0cathode-E0anode #standard emf of the cell, V\n",
- "import math\n",
- "K=math.exp(round(n*E0cell/0.0257,1)) #equilibrium constant, from the formula E0cell=0.0257*lnK/n\n",
- "\n",
- "#Result\n",
- "print\"The equilibrium constant for the given reaction is :%.e\"%K "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.29\n",
- "The equilibrium constant for the given reaction is :7e+09\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:19.5,Page no:851"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "n=6 \n",
- "F=96500 #faraday constant, J/V mol\n",
- "E0cathode=-2.87 #standard electrode potential of cathode(Ca2+/Ca), V\n",
- "E0anode=1.5 #standard electrode potential of anode(Au3+/Au), V\n",
- "\n",
- "#Calculation\n",
- "E0cell=E0cathode-E0anode #standard emf of the cell, V\n",
- "deltaG0=-n*F*E0cell #standard free energy change for the reaction, kJ/mol\n",
- "\n",
- "#Result\n",
- "print\"The standard free energy change for the reaction is :%.2e\"%(deltaG0/1000),\"kJ/mol\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The standard free energy change for the reaction is :2.53e+03 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:19.6,Page no:853"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "n=2 \n",
- "F=96500 #faraday constant, J/V mol\n",
- "Co2=0.15 #conc of Co2+ ions, M\n",
- "Fe2=0.68 #conc of Fe2+ ions, M\n",
- "E0cathode=-0.44 #standard electrode potential of cathode(Fe2+/Fe), V\n",
- "E0anode=-0.28 #standard electrode potential of anode(Co2+/Co), V\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "E0cell=E0cathode-E0anode #standard emf of the cell, V\n",
- "Ecell=E0cell-0.0257/2*math.log(Co2/Fe2) #calculation of cell potential at non standard conditions, V\n",
- "\n",
- "#Result\n",
- "print \"E=\",round(Ecell,2),\"V\"\n",
- "if(Ecell>0):\n",
- " print\"The reaction would proceed spontaneously in the direction written\"\n",
- "else:\n",
- " print\"The reaction is not spontaneously in the direction written\"\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "E= -0.14 V\n",
- "The reaction is not spontaneously in the direction written\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:19.7,Page no:855"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "n=2 \n",
- "Zn=1 #conc of Zn2+ ions, M\n",
- "pH2=1 #pressure of H2 gas, atm\n",
- "Ecell=0.54 #emf of the cell, V\n",
- "E0cell=0.76 #standard emf of the cell, V\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Q=math.exp(-(Ecell-E0cell)*2/0.0257) #since Ecell=E0cell-0.0257/2*log(Q) where Q=(Zn2+)*pH2/(H+)**2\n",
- "H=math.sqrt(Zn*pH2/Q) #the conc of H+ ions, M\n",
- "\n",
- "#Result\n",
- "print\"The molar concentration of H+ ion is :%.e\"%H,\"M\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The molar concentration of H+ ion is :2e-04 M\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:19.9,Page no:870"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "t=7.44*3600 #time, sec\n",
- "A=1.26 #current in ampere\n",
- "F=96500.0 #faraday constant, J/V mol\n",
- "R=0.0821 #gas constant, L atm/K\n",
- "T=273.0 #temperature in Kelvin\n",
- "P=1.0 #pressure in atm\n",
- "\n",
- "#Calculation\n",
- "q=A*t #charge passed, coulomb\n",
- "ne=q/F #moles of electrons\n",
- "nO2=ne/4.0 #moles of oxygen\n",
- "nH2=ne/2.0 #moles of H2\n",
- "VO2=nO2*R*T/P #volume of oxygen gas generated\n",
- "VH2=nH2*R*T/P #volume of H2 gas generated\n",
- "\n",
- "#Result\n",
- "print\"The volume of O2 gas is:\",round(VO2,2),\"L\"\n",
- "print\"Volume of H2 gas generated is:\",round(VH2,2),\"L\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The volume of O2 gas is: 1.96 L\n",
- "Volume of H2 gas generated is: 3.92 L\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
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