How can I do redox reactions step by step?

1 Answer
Jul 1, 2018

Answer:

A rather open-ended question...

Explanation:

And the first step is assignment of oxidation state...for elemental ions, i.e. #Na^+#, or #K^+# or #Cr^(3+)#, the oxidation state of the metal is the CHARGE on the ion, so that we gots #Na(+I), K(+I), Cr(+III)# respectively...

For compound ions, we have to divide up the anion/cation according to predetermined rules

Now #"permanganate"#, #MnO_4^(-)#, i.e. #Mn(+VII)# is a brightly coloured anion, that commonly undergoes reduction to give COLOURLESS #Mn^(2+)#..and since #+VII-II=V#... WE CONCEIVE that manganese undergoes a FIVE ELECTRON REDUCTION....

#underbrace(MnO_4^(-))_"deep purple"+8H^+ +5e^(-) rarr Mn^(2+)+4H_2O# #(i)#

And note for this half equation both mass and charge are balanced, as is absolutely required. Now the electrons are conceived to COME from somewhere...to cause a corresponding oxidation...let us assume that it OXIDIZES a primary alcohol as shown...i.e. here we gots the oxidation #C(-I)rarrC(+III)#, a four electron oxidation....

#Rstackrel(-I)CH_2OH+H_2O(l)rarrRstackrel(+III)CO_2H+4H^+ +4e^(-)# #(ii)#

We add #(i)# and #(ii)# in such a way that the electrons, virtual particles of convenience, are RETIRED....and thus #4xx(i)+5xx(ii)# gives....

#4MnO_4^(-)+32H^+ +5RCH_2OH+5H_2O(l) +20e^(-) rarr 4Mn^(2+)+16H_2O+5RCO_2H+20H^+ +20e^(-)#

To give after cancellation...

#4MnO_4^(-) +5RCH_2OH +12H^+rarr 4Mn^(2+)+5RCO_2H+11H_2O#

The which is balanced with respect to mass and charge as is absolutely required...

Confused yet? But all I have done is assign oxidation numbers according to predetermined rules...and based electron transfer on the basis of the change in oxidation number....