REDOX & Electrochemistry - LSU Chemistry
REDOX 22Example: What is the potential for the followingreaction?2Na + (aq) + 2Cl − (aq)Cl 2 (g) + 2Na(s)Reduction half cell: Na + (aq) + 1e- Na(s) −2.71 VOxidation half cell: 2Cl − (aq) Cl 2 (g) + 2e- −1.36 Vsum = −4.07 VThis half-cell was theone that was flippedaround (reversed)Non-spontaneousThe reverse reaction is very spontaneous with a cellpotential of +4.07 V.
Example: What is the potential for theelectrochemical cell composed of the following 2When phrased like this, we are looking for a spontaneoushalf cell rxns? reaction that gives a positive cell potential!Cr +3 (aq) + 3e- Cr(s) -0.74MnO 2 + 4H + + 2e - Mn 2+ (aq) + 2H 2 O +1.28The Cr +3 (aq) + 3e- → Cr(s) rxn has the morenegative potential, so it gets flipped around:Cr(s) Cr +3 (aq) + 3e- +0.74MnO 2 + 4H + + 2e - Mn 2+ (aq) + 2H 2 O +1.28REDOX 23Now we have to balance the # of electrons in each rxn so wecan add them together. 6 e- is the common factor, so we needto multiply the Cr rxn by 2 and the MnO 2 rxn by 3:2[Cr(s) Cr +3 (aq) + 3e- ] +0.743[MnO 2 + 4H + + 2e - Mn 2+ (aq) + 2H 2 O] +1.282Cr(s) + 3MnO 2 + 12H +2Cr +3 (aq) + 3Mn 2+ (aq) + 6H 2 O +2.02VNote that you do NOT multiply the cellpotentials by the numerical factors (2or 3) used to balance the # of electronsin each half-cell rxn!!DANGER!!Commonmistake!!