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diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_6.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_6.ipynb new file mode 100644 index 00000000..b15856da --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_6.ipynb @@ -0,0 +1,620 @@ +{ + "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": {} + } + ] +}
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