Half Equations

Write half equations for the displacement reaction between bromine and iodide ions. The unbalanced equation is given below:

\(Br_2+I^-\rightarrow I_2+Br^- \)

First, let’s pick a reactant. We'll start with bromine. Bromine (Br₂) reacts to form bromide ions (Br^-):

\(Br_2\rightarrow Br^- \)

This equation isn’t balanced. There are two Br on the left, but only one on the right. To balance it, we need to double the number of bromide ions:

\(Br_2\rightarrow 2Br^- \)

\(Br_2+2e^-\rightarrow 2Br^- \)

Both sides now have the same number of each element and the same charge. The half equation is balanced. But we’re not done. We now need to write a second half equation for iodine.

In this reaction, iodide ions react to form iodine:

\(I^-\rightarrow I_2\)

Balancing the equation in terms of I gives us the following:

\(2I^-\rightarrow I_2\)

This time, the charge on the left-hand side of the equation is 2(-1) = -2, and the charge on the right-hand side is +0. This means that we need to add two negative electrons to the right-hand side:

\(2I^-\rightarrow I_2+2e^- \)

Once again, this equation is now balanced in terms of elements and charges. Your two half equations are complete.

Write half equations for the reaction between manganate(VII) ions and iron(II) ions to form manganese(II) ions and iron(III) ions. The unbalanced equation is given below:

\(MnO_4^- + Fe^{2+} + H^+ \rightarrow Mn^{2+} + Fe^{3+} + H_2O\)

To balance the charges, we need to add one electron to the right-hand side of the equation:

\(Fe^{2+} \rightarrow Fe^{3+} + e^-\)

Both elements and charges are balanced; that's our first half equation done. We'll now consider the manganate ion. Here's our starting equation:

\(MnO_4^- \rightarrow Mn^{2+}\)

There are the same number of Mn on each side of the equation, so we don’t need to worry about them. However, there are four oxygens (O) on the left-hand side but none on the right-hand side. We need to balance the equation by adding more O to the right-hand side. Remember that the only substances we can add to half equations are water (H₂O), hydrogen ions (H⁺), and electrons (e^-). So, to add more oxygen to the right-hand side, we need to include some H₂O. We need 4 O, so we add in 4 H₂O:

\(MnO_4^- \rightarrow Mn^{2+} + 4H_2O\)

We've now encountered another issue: there are 8 hydrogens (H) on the right-hand side of the equation, but none on the left. Luckily for us, hydrogen ions (H⁺) are one of the species that we are allowed to add to half equations. Therefore, we add 8 H⁺ to the left-hand side:

\(MnO_4^- + 8H^+\rightarrow Mn^{2+} + 4H_2O\)

We're almost done. However, the charges aren't balanced: There is an overall charge of +7 on the left-hand side, but only +2 on the right. To make the charges equal, we add 5 negative electrons to the left-hand side:

\(MnO_4^- + 8H^+ + 5e^-\rightarrow Mn^{2+} + 4H_2O\)

Give the equation a final check to make sure that the number of moles of each element and the overall charge is balanced on each side of the equation. In this case, everything looks good. Well done - we've written our two half equations.

Combine the following two half equations to create one overall redox equation:

\(Fe^{2+} \rightarrow Fe^{3+} + e^-\)

\(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\)

Take a look at the two half equations. The first has just one electron on the right-hand side, whilst the second has five on the left-hand side. We need to multiply each of the half equations by a constant so that they both feature the same number of electrons. The easiest way to do this is by multiplying the first equation by 5. This way, both equations will feature five electrons:

\(5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-\)

We now add the two reactions together to form one overall redox equation. Add all the reactants from both half equations to the left-hand side of this new redox equation, and add all the products to the right:

\(5Fe^{2+} + MnO_4^- + 8H^+ + 5e^- \rightarrow 5Fe^{3+} + 5e^-+ Mn^{2+} + 4H_2O\)

You'll see that the electrons cancel out. We can remove these, giving us our final answer:

\(5Fe^{2+} + MnO_4^- + 8H^+ +\cancel {5e^-}\rightarrow 5Fe^{3+} + \cancel {5e^-}+ Mn^{2+} + 4H_2O\)

\(5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O\)