## zhiyuan3yu5 3 years ago potassium benzoate, C6H5CO2K, can be prepared from benzyl alcohol, C6H5CH2OH and KMnO4 in aqueous KOH. The manganese is converted to MnO2. balance the equation. I know that redox is used to solve this. can someone just get me started on what i need to do to set up the half reactions. i'm taking orgo and it's been a while... 3C6H5CH2OH+4KMnO4--->3C6H5CO2K+4H2O+KOH+4MNO2

1. Carl_Pham

First write the net ionic equation, so you get rid of the spectator ions: ${\rm C}_6{\rm H}_5{\rm CH}_2{\rm OH}(aq) + {\rm MnO}_{4}^{-}(aq) \rightarrow ({\rm C}_6{\rm H}_5{\rm CO}_2)^{-}(aq) + {\rm MnO}_2(s)$ Next find out what's oxidized and what's reduced by assigning oxidation numbers. You'll find the only changes are that the Mn goes from +7 to +4 (so it's reduced) and the primary carbon goes from 0 in the alcohol to +4 in the carboxylate (so it's oxidized). Now write the redox half reactions by writing ONLY the reactants and products that contain the oxidized and reduced atom, e.g. for the reduction half reaction write ONLY the reactants and products that contain the Mn: ${\rm MnO}_{4}^{-}(aq) \rightarrow {\rm MnO}_2(s)$ Since the Mn is reduced from +7 to +4, it must gain three electrons: ${\rm MnO}_{4}^{-}(aq) + 3 e ^- \rightarrow {\rm MnO}_2(s)$ Balance oxygen by adding water: ${\rm MnO}_{4}^{-}(aq) + 3 e^- \rightarrow {\rm MnO}_2(s) + 2 {\rm H}_2{\rm O}(l)$ Balance hydrogen by adding H+: ${\rm MnO}_{4}^{-}(aq) + 4 {\rm H}^{+}(aq) + 3 e^- \rightarrow {\rm MnO}_2(s) + 2 {\rm H}_2{\rm O}(l)$ So far, you've balanced as if you're in acidic solution, but you're actually in basic, so correct by adding OH- to both sides (so the atoms stay balanced) until all the H+ is gone, combining H+ and OH- that appear on the same side into H2O: ${\rm MnO}_{4}^{-}(aq) + 4 {\rm H}_2{\rm O}(l) + 3 e^- \rightarrow {\rm MnO}_2(s) + 2 {\rm H}_2{\rm O}(l) + 4 {\rm OH}^{-}(aq)$ Cancel any waters that appear on both side: ${\rm MnO}_{4}^{-}(aq) + 2 {\rm H}_2{\rm O}(l) + 3 e^- \rightarrow {\rm MnO}_2(s) + 4 {\rm OH}^{-}(aq)$ Double check that your charge is balanced -- it is, -4 on both sides -- and you've got your reduction half reaction. Do the same for the oxidation half reaction. Now multiply the oxidation half reaction by the numbe of electrons in the reduction half reaction, and the reduction half reaction by the number of electrons in the oxidation half reaction. Add. All the electrons should cancel. Cancel any other compounds that appear on both sides. Divide stoichiometric coefficients by any common factors, e.g. if they're all divisible by 2, divide through by 2.

2. zhiyuan3yu5

i did the oxidation half reaction, and got C6H5CH2OH+5OH- ---> C6H5CO2- + 4H2O+ 4e- and then multiplied the reduction reaction by 4 and the oxidation reaction by 3. after cancelling things out i got 3C6H5CH2OH +4MnO4- ----> 4MnO2+3C6H5CO2- + OH- + 4H2O so do i just add the K now to 4MnO4- , 3C6H5CO2 and OH- ? and if so why do we do that at this point?

3. waheguru
4. waheguru