# Question 6dd0a

Aug 14, 2016

The $n$ factor for redox reactions, specifically, is just the number of electrons transferred in the redox reaction for each $\text{mol}$ 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 $\boldsymbol{\text{H"_2"O}}$, 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 $\text{H"_2"O}$) by adding $\boldsymbol{{\text{H}}^{+}}$, 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 ${\text{MnO}}_{4}^{-}$ to ${\text{Mn}}^{2 +}$ (${E}_{\text{red"^@ = +"1.49 V}}$) in strong acid. We can use the oxidation of $\text{Fe} \left(s\right)$ to ${\text{Fe}}^{2 +} \left(a q\right)$ (${E}_{\text{ox"^@ = +"0.44 V}}$).

REDUCTION

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

$\stackrel{\textcolor{red}{+ 7}}{{\text{Mn")stackrel(color(red)(-2))("O"_4^(-))(aq) -> stackrel(color(red)(+2))("Mn}}^{2 +} \left(a q\right)}$

The $\text{Mn}$ are balanced, so we move to oxygen.

$\text{MnO"_4^(-)(aq) -> "Mn"^(2+)(aq) + 4"H"_2"O} \left(l\right)$

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

$\text{MnO"_4^(-)(aq) + 8"H"^(+)(aq) -> "Mn"^(2+)(aq) + 4"H"_2"O} \left(l\right)$

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 $\boldsymbol{5}$ electrons to the left side balances out the charge as $+ 2$ on each side.

That means our "n-factor" is $\boldsymbol{5}$ for $\text{Mn}$ in this half-reaction. It doesn't mean it will be $5$ for $\text{Mn}$ species in any half-reaction.

$\textcolor{g r e e n}{\text{MnO"_4^(-)(aq) + 8"H"^(+)(aq) + 5e^(-) -> "Mn"^(2+)(aq) + 4"H"_2"O} \left(l\right)}$

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.

$\textcolor{g r e e n}{\stackrel{\textcolor{red}{0}}{{\text{Fe"(s)) -> stackrel(color(red)(+2))("Fe}}^{2 +}} \left(a q\right) + 2 {e}^{-}}$

That means our "n-factor" is $\boldsymbol{2}$ for $\text{Fe}$ in this half-reaction. It doesn't mean it will be $2$ for $\text{Fe}$ 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.

$2 \left(\text{MnO"_4^(-)(aq) + 8"H"^(+)(aq) + cancel(5e^(-)) -> "Mn"^(2+)(aq) + 4"H"_2"O} \left(l\right)\right)$
5("Fe"(s) -> "Fe"^(2+)(aq) + cancel(2e^(-))#)
$\text{---------------------------------------------------}$
$\textcolor{b l u e}{2 \text{MnO"_4^(-)(aq) + 5"Fe"(s) + 16"H"^(+)(aq) -> 2"Mn"^(2+)(aq) + 5"Fe"^(2+)(aq) + 8"H"_2"O} \left(l\right)}$