Question

Look up the reduction potential for ${\mathrm{Fe}}^{3+}$ to ${\mathrm{Fe}}^{2+}$. Look up the reduction potential for ${\mathrm{Fe}}^{2+}$ to Fe. Finally, look up the reduction potential for ${\mathrm{Fe}}^{3+}$ to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two potentials to calculate the third potential.

Short answer

The sum of the reduction potentials for the first two reactions does not equal the potential for the third reaction because the stoichiometry of the electrons involved in the reactions is not the same. To find the reduction potential for the third reaction, balance the electrons in the first reaction and then add it to the second reaction. Calculating the potential for the third reaction using the first two potentials, we find that $E_3^{\circ }=1.10V$.

Step-by-step solution

Look up the reduction potentials for the given reactions

First, we need to look up the reduction potentials for the three reactions. We can find these values in a table of reduction potentials or in a textbook. (1) ${\mathrm{Fe}}^{3+}+e^{-}\to {\mathrm{Fe}}^{2+}$, $E_1^{\circ }=0.77V$ (positive value means spontaneous reduction) (2) ${\mathrm{Fe}}^{2+}+2e^{-}\to \mathrm{Fe}$, $E_2^{\circ }=-0.44V$ (negative value means non-spontaneous reduction) (3) ${\mathrm{Fe}}^{3+}+3e^{-}\to \mathrm{Fe}$, $E_3^{\circ }=?$

Explain why adding reduction potentials is not valid for the third reaction

Adding the reduction potentials for the first two reactions does not give the potential for the third because the stoichiometry of the electrons involved is not the same for all reactions. In the first reaction, only 1 electron is involved, while in the second reaction, 2 electrons are involved. However, in the third reaction, 3 electrons are involved. The stoichiometry of the electrons has to be taken into account when summing up reduction potentials to obtain a new potential.

Use the first two potentials to calculate the third potential

In order to use the first two potentials to calculate the potential of the third reaction, we have to manipulate the reactions in such a way that when they are added together, the desired balanced equation for the third reaction is obtained. First, we multiply the first reaction by 2 to balance the electrons: (1) $2({\mathrm{Fe}}^{3+}+e^{-}\to {\mathrm{Fe}}^{2+})\Rightarrow 2{\mathrm{Fe}}^{3+}+2e^{-}\to 2{\mathrm{Fe}}^{2+}$, $2E_1^{\circ }=1.54V$ Then, we add up the modified first reaction and the second reaction: $2{\mathrm{Fe}}^{3+}+2e^{-}\to 2{\mathrm{Fe}}^{2+}$ ${\mathrm{Fe}}^{2+}+2e^{-}\to \mathrm{Fe}$ $+\Downarrow$ $2{\mathrm{Fe}}^{3+}+3e^{-}\to \mathrm{Fe}+{\mathrm{Fe}}^{2+}$ This gives us the desired third reaction with one more iron(II) product on the right side. Now, we can sum up their reduction potentials: $E_3^{\circ }=2E_1^{\circ }+E_2^{\circ }=1.54V-0.44V=1.10V$ So, the reduction potential for the third reaction is $E_3^{\circ }=1.10V$.