How can the iron in the reaction of iron (ii) chloride and zinc not go to zero oxidation state?

Question

I know that the zinc half equation would be $\ce{Zn -> Zn^{2+} + 2e-}$.

However I am confused about the half equation for iron. I would think that since iron is completely displaced, its oxidation state goes to zero, so its half equation would be $\ce{Fe^{2+} + 2e- -> Fe}$.

However my teacher said that whenever you have iron, it will always change oxidation states from 2+ to 3+ or vice-versa in a redox reaction, and will not go to zero. For instance, she gave us an example in which there was a reaction between iron(iii) chloride and zinc. I thought that the zinc would completely displace the iron so that it goes from 3+ ion to 0 in the solid iron metal. However my teacher insisted that it goes to 2+, and not to zero.

In the example of iron(ii) chloride, I don't even see how it is possible for it to go to 3+ because then both zinc and iron are oxidised!

Answer 1

Generally, to predict redox reactions, we consider the Standard Reduction Potentials at $25\ ^\circ\mathrm{C}$: The bigger the difference between the potentials of two redox couples, more favorable is the reaction. $$E^0(\ce{Fe^3+ /Fe^2+})=0.77\ \mathrm{V}$$ $$E^0(\ce{Fe^3+ /Fe})=-0.04\ \mathrm{V}$$ $$E^0(\ce{Fe^2+ /Fe})=-0.44\ \mathrm{V}$$ $$E^0(\ce{Zn^2+ /Zn})=-0.76\ \mathrm{V}$$ So, when $\ce{Fe^3+}$ reacts with $\ce{Zn}$, we get $\ce{Fe^2+}$ (not $\ce{Fe}$) and $\ce{Zn^2+}$.

As for your reaction: when $\ce{Fe^2+}$ reacts with $\ce{Zn}$, we get $\ce{Fe}$ and $\ce{Zn^2+}$.