Calculate the percentage difference in the fundamental vibrational wavenumbers of \({ }^{23} \mathrm{Na}^{35} \mathrm{Cl}\) and \({ }^{23} \mathrm{Na}^{37} \mathrm{Cl}\) on the assumption that their force constants are the same.
Text Transcription:
^23Na^35Cl
^23Na^37Cl
Reference Electrode There is no way to measure reduction potential of an isolated half reaction Only the difference is potential between 2 half cells can be added In order to assign E nought red to half reactions, reference electrode chosen, where the cell potential = 0.0 V Standard H Electrode 2H++2e>H2 EH+=0.0 V reversible not at equilibrium can be oxidation or reduction Example 5: Galvanic Cell Zn/Zn2+ Zn>Zn2++2e anode 2H++2e_>H2 cathoden+2H+>Zn2++H2 Ecell= ErEo or Ecell= EcEa 0.76=0VEZn2+=0.76 EZn2+ Ered table 20.1 in the book -oxidized+e→ reduced ions elements or compounds Top to bottom down E nought red low value Ered<0, easily oxidized Value Ered=less likely to undergo reduction Sign Ered attached to H+/H2 E red>0, easily reduced to undergo reduction F+E2>2F E=2.87 (most easily reduced) Li++e>Li Erednought= 3.05, reverse reaction will occur Using E cell to Calculate E red galvanic cell 2Ag++Cu>2Ag+2Cu2+ Ecell=0.46 V cu 0.34 V Ag+ reduction Ecell= EAg+ECu 0.46 V=EAg+(0.34) =EAg+=0.80V Fixed Spontaneity 2 half reactions Ecell >0 delta G<0 reaction is spontaneous More + reduction occurs as written (reduction Less + Enought red forced to go in reverse oxidation E cell 1st Half reaction more + goes up reduction, other half decreases oxidation Left + half reaction, right half reaction Example 6: Cr+Au>Au3++Cr3+ 1.5+0.74=2.24 V Cr3++3e
