Question #6dd0a

The #n# factor for redox reactions, specifically, is just the number of electrons transferred in the redox reaction for each #"mol"# of a given substance.

NOTE: If a substance contains two reactants, one of which is reduced and the other of which is oxidized, then their individual n-factors sum to be the n-factor of the substance at hand.

I find it easiest if you find the number of electrons last, because then you can use them to balance the charge, and that tells you the "n-factor" for that element.

Here's what I tend to do:

1. Balance the major elements by adding stoichiometric coefficients, if needed (e.g. what is being reduced/oxidized), in each half-reaction.
2. Balance the oxygen atoms by adding #bb("H"_2"O")#, if needed, in each half-reaction. That is, we assume our reaction is in aqueous solution.
3. Balance the hydrogen atoms (some were added in writing #"H"_2"O"#) by adding #bb("H"^(+))#, in each half-reaction.
4. Balance the charge with electrons, for each half-reaction.

So a random example. Let's look at the reduction of #"MnO"_4^(-)# to #"Mn"^(2+)# (#E_"red"^@ = +"1.49 V"#) in strong acid. We can use the oxidation of #"Fe"(s)# to #"Fe"^(2+)(aq)# (#E_"ox"^@ = +"0.44 V"#).

REDUCTION

The basic thing to start with are the starting and ending species.

#stackrel(color(red)(+7))("Mn")stackrel(color(red)(-2))("O"_4^(-))(aq) -> stackrel(color(red)(+2))("Mn"^(2+)(aq))#

The #"Mn"# are balanced, so we move to oxygen.

#"MnO"_4^(-)(aq) -> "Mn"^(2+)(aq) + 4"H"_2"O"(l)#

Now the protons to balance out the hydrogens we added due to water (we stated that we had strong acid).

#"MnO"_4^(-)(aq) + 8"H"^(+)(aq) -> "Mn"^(2+)(aq) + 4"H"_2"O"(l)#

And at this point we would add the electrons to balance out the net charge. The left has a charge of #-1 + 8 = +7#, and the right side has a net charge of #+2#.

Therefore, adding #bb(5)# electrons to the left side balances out the charge as #+2# on each side.

That means our "n-factor" is #bb(5)# for #"Mn"# in this half-reaction. It doesn't mean it will be #5# for #"Mn"# species in any half-reaction.

#color(green)("MnO"_4^(-)(aq) + 8"H"^(+)(aq) + 5e^(-) -> "Mn"^(2+)(aq) + 4"H"_2"O"(l))#

OXIDATION

This one is pretty simple. There are no oxygens, no hydrogens, nothing else to balance other than the charge.

So just add #2# electrons to make both sides have a net neutral charge.

#color(green)(stackrel(color(red)(0))("Fe"(s)) -> stackrel(color(red)(+2))("Fe"^(2+))(aq) + 2e^(-))#

That means our "n-factor" is #bb(2)# for #"Fe"# in this half-reaction. It doesn't mean it will be #2# for #"Fe"# species in any half-reaction.

OVERALL REACTION

And now let's finish up this reaction. Make sure the electrons cancel out by scaling each half-reaction.

#2("MnO"_4^(-)(aq) + 8"H"^(+)(aq) + cancel(5e^(-)) -> "Mn"^(2+)(aq) + 4"H"_2"O"(l))#
#5("Fe"(s) -> "Fe"^(2+)(aq) + cancel(2e^(-))#)
#"---------------------------------------------------"#
#color(blue)(2"MnO"_4^(-)(aq) + 5"Fe"(s) + 16"H"^(+)(aq) -> 2"Mn"^(2+)(aq) + 5"Fe"^(2+)(aq) + 8"H"_2"O"(l))#