Electrons in Nernst equation depend on how it is balanced

Question

Using the following equation: $K = \mathrm{e}^{zFE/RT}=10^{zE/0.059}$ (standard conditions) and the following disproportionation reaction:
\begin{align} \ce{BrO- + 2e- + H2O &-> Br- + 2OH-}\\ \frac{1}{2}(\ce{BrO- + 4OH- &-> 2H2O + BrO3- + 4e-})\\\hline \ce{\frac{3}{2} BrO- &-> \frac{1}{2}BrO3- + Br-} \end{align}

is $Z=2$ or $4$ (professor multiplied the first reaction by 2)? The electrode potential does not depend on the number of transferred electrons but how (and why) do we determine the correct number of electrons for this case?

Answer 1

You match the number of electrons with the expression for $K$ itself, which in turn depends on the coefficients you use to balance the equation. If you do everything right you get the same algebraic relation that reflects the correct thermodynamics.

Thus for hypobromite disproportionation we may have

$\ce{(3/2)BrO^- <=> (1/2)BrO3^- + Br^-}$

with two electrons transferred due to one bromine atom being reduced from $+1$ to $-1$ oxidation state. Therefore,

$K=\color{blue}{\dfrac{[\ce{BrO3^-}]^{1/2}[\ce{Br^-}]}{[\ce{BrO^-}]^{3/2}}=10^{2E/(0.059\text{ V})}}$

Or, we can render it as

$\ce{3BrO^- <=> BrO3^- + 2Br^-}$

with four electrons transferred due to two bromine atoms being reduced from $+1$ to $-1$ oxidation state. Therefore,

$K=\color{blue}{\dfrac{[\ce{BrO3^-}][\ce{Br^-}]^2}{[\ce{BrO^-}]^3}=10^{4E/(0.059\text{ V})}}$

The two formulations will give different $K$ values. But the blue parts of the equations are equivalent; the second equation is just the first one with both sides squared. So if you properly formulate $K$ based on the coefficients along with the electron count, you get the same empirically accessible relationships between the reactant and product concentrations/activities.