When we scale a half reaction in order to balance the number of electrons with the other half reaction, we do not change the intrinsic joules per coulomb (Volts) of the electrons.
So Fe2+ +2e- -------> Fe (s) E = -0.44V
means that the 2 electrons carry -0.44 Joules per coulomb of energy
If we double the reaction, the reaction energy is doubled, but so is the number of electrons. So the J/C stays the same. That is the standard reduction potential remains -0.44V
So 2Fe2+ +4e- -------> Fe (s) E = -0.44V
But when we are trying to combine the 2 reduction reactions as in the case of:
Fe2+ +2e- -------> Fe (s) E = -0.44V
Fe3+ +e- -------> Fe2+ (s) E = +0.77V
to get the standard reduction potential for :
Fe3+ +3e- -------> Fe (s) E = ??
this is a different concept. Here we are not scaling half reactions in order to combine them into a single full reaction. We are trying to combine 2 half reactions to make a new half reaction (with electrons still present in the equation)
At first glance it's tempting to just add the potentials, since adding the half reactions does give the desired half reaction :
- Fe2+ +2e- -------> Fe (s) E = -0.44V
- Fe3+ +e- -------> Fe2+ (s) E = +0.77V
Fe2+ +2e- Fe3+ +e- -------> Fe (s) Fe2+ (s) E = ?
(now cancel the Fe2+ on each side)
- Fe3+ +3e- -------> Fe (s) E = ?
But we cant just add the potentials. This is still a half reaction, so the potential is all about the energy per electron in Joules /Coulomb (V)
In reaction 1) the 2 electron carry -0.44 Joules per coulomb, and in reaction 2) the 1 electron carries 0.77 Joules per coulomb
So to get the J/C for reaction 3) we need the weighted average J/C of the 3 electrons
So that 2/3 electrons at -0.44 plus 1/3 electrons at +0.77
2/3 * (-0.44) + 1/3 * (0.77) = =0.037 Joules per coulomb (volts)
Hope that helps.