Half equations  Redox and Electrochemistry
Half equations
Lessons
Notes:
In this lesson, we will learn:
 How to complete and balance half equations in basic and acidic conditions.
 How to use half equations to balance full redox equations.
Notes:
 Because redox reactions always involve both reduction and oxidation, to understand them more easily the ‘whole’ reaction can be split into two ‘halfreactions’ or halfequations. Halfequations are the two reduction and oxidation processes written separately, as if they were occurring alone.
Redox exam questions normally ask you to complete a redox equation, but only give you the major elements (anything except O and H) to start. They may or may not already be split into halfequations. Either way, you need to add everything else to complete it.
This lesson will show you how to balance halfequations so that they are ready to be combined for the full redox equation. There is one worked example for this here; more practice on this last part is in Balancing redox equations.  The most straightforward way to do this is to split a redox reaction into halfequations and complete these steps in order, using a worked example:
MnO_{4}^{} $\enspace$→$\enspace$ Mn^{2+}
 First, balance the major elements. This is any atom that isn’t oxygen or hydrogen. There is normally only one input reactant and output product for these, so just increase the number of moles to balance.
In our example, manganese is already balanced so no change is necessary:MnO_{4}^{} $\enspace$→$\enspace$ Mn^{2+}  Next, balance oxygen. Do this by adding H_{2}O molecules (remember redox reactions nearly always occur in aqueous solution!).
There are currently four oxygen atoms in the reactants and none in the products, so we add four H_{2}O molecules there:MnO_{4}^{} $\enspace$→$\enspace$ Mn^{2+} + 4H_{2}O  Next, balance hydrogen. This is done by adding H^{+ }_{
(aq)} ions.
There are currently eight hydrogens in the products and none in the reactants, so we add eight H^{+} there:MnO_{4}^{} + 8H^{+} $\enspace$→$\enspace$ Mn^{2+} + 4H_{2}O  Finish balancing charge by adding electrons (e^{}).
The reactants have an overall +7 charge while the products have an overall +2 charge, so 5 electrons (a total of 5 charge) needs to be ‘added’ to the reactants to bring it down to +2:MnO_{4}^{} + 8H^{+} + 5e^{} $\enspace$→$\enspace$ Mn^{2+} + 4H_{2}O
This is the complete half equation and is accurate for acidic conditions. As always, here you can manually check the number of atoms and charge on each side to be sure.
If the reaction is in basic conditions then you must make a change because H^{+} will not be present, OH^{} will be. To balance a half equation in basic conditions from here, you need to use the autoionization of water equilibrium:H^{+} + OH^{} $\enspace \rightleftharpoons \enspace$ H_{2}O
Multiply this equation by the number of H^{+} in your balanced half equation and insert it so that H^{+} from the halfequation and H^{+} from the autoionization of water equilibrium will be on opposite sides. This will cancel the terms out and leave you with just H_{2}O and OH^{} in the halfequation, which is correct for basic conditions!
Our example (the water equilibrium compounds are highlighted in red ) looks like this:MnO_{4}^{} + 8H^{+} + 5e^{} + 8H_{2}O $\enspace$→$\enspace$ Mn^{2+} + 4H_{2}O + 8H^{+} + 8OH^{}
This leaves us with equal H^{+} and some H_{2}O on both sides of the equation. Cancel it out:MnO_{4}^{} + 8H^{+}+ 5e^{} + 8 4H_{2}O $\enspace$→$\enspace$ Mn^{2+} +4H_{2}O+8H^{+}+ 8OH^{}
This gives us the correct half equation in basic conditions:MnO_{4}^{} + 5e^{} + 4H_{2}O $\enspace$→$\enspace$ Mn^{2+} + 8OH^{}  First, balance the major elements. This is any atom that isn’t oxygen or hydrogen. There is normally only one input reactant and output product for these, so just increase the number of moles to balance.
 WORKED EXAMPLE 2: Balance in basic conditions.
MnO_{2} $\enspace$→$\enspace$ Mn^{2+}  First, balance major atoms:
MnO_{2} $\enspace$→$\enspace$ Mn^{2+}  Next, balance oxygen.
MnO_{2} $\enspace$→$\enspace$ Mn^{2+} + 2H_{2}O  Next, balance hydrogen.
MnO_{2} + 4H^{+}$\enspace$→$\enspace$ Mn^{2+} + 2H_{2}O  Balance by adding e^{}.
MnO_{2} + 4H^{+} + 2e^{}$\enspace$→$\enspace$ Mn^{2+} + 2H_{2}O  As this is in basic conditions, again we need to cancel out the H^{+} which will not be present – we need to balance this using the dissociation of water equation.
H^{+} + OH^{} $\enspace \rightleftharpoons \enspace$ H_{2}O
There are four H^{+} in our current halfequation:MnO_{2} + 4H^{+} + 2e^{}$\enspace$→$\enspace$ Mn^{2+} + 2H_{2}O
We therefore need to multiply the water equation four times and insert it in to our halfequation to cancel out the four H^{+} already present.MnO_{2} + 4H^{+} + 4H_{2}O + 2e^{}$\enspace$→$\enspace$ Mn^{2+} + 2H_{2}O + 4H^{+} + 4OH^{}
This allows us to cancel the H^{+} that was in our half equation already and the H^{+} from the water equation we just added.MnO_{2} + 4H^{+}+ 4H_{2}O + 2e^{}$\enspace$→$\enspace$ Mn^{2+} +2H_{2}O+4H^{+}+ 4OH^{}
This gives a final equation:MnO_{2} + 2H_{2}O + 2e^{}$\enspace$→$\enspace$ Mn^{2+} + 4OH^{}
 First, balance major atoms:
 WORKED EXAMPLE 3: With the ability to build correct halfequations, we can now combine two halfequations and make full equations. This is done by checking the number of electrons in each halfequation.
Like mass, charge is conserved in a reaction, so the number of electrons in the reactants must equal the number in the products!
To combine halfequations into a full equation, multiply the half equations until you have equal electrons on both sides, then combine them. Using an example, with two half equations:MnO_{4}^{} + 8H^{+} + 5e^{}$\enspace$→$\enspace$ Mn^{2+} + 4H_{2}O $\qquad \qquad \qquad \enspace \enspace$ Os + 4H_{2}O $\enspace$→$\enspace$ OsO_{4} + 8H^{+} + 8e^{} $\quad$ (acidic solution)
Remember that electrons lost equal electrons gained in a redox reaction, so we must have an equal amount on both sides. To get this, we need to multiply the equations until we get a common number – in our example, we have to find a number that both 5 and 8 go into!$\qquad$MnO_{4}^{} + 8H^{+} + 5e^{} $\enspace$→$\enspace$ Mn^{2+} + 4H_{2}O $\qquad$ x8 $\qquad \quad$Os + 4H_{2}O $\enspace$→$\enspace$ OsO_{4} + 8H^{+} + 8e^{} $\qquad \enspace$ x5
Multiplying by these gives us 40 electrons in each half equation – these will now cancel when we combine the equation:8MnO_{4}^{} + 64H^{+} + 40e^{} + 5Os + 20H_{2}O $\enspace$→$\enspace$ 8Mn^{2+} + 32H_{2}O + 5OsO_{4} + 40H^{+} + 40e^{}
Cancel the common species:8MnO_{4}^{} + 6424H^{+} +40e^{}+ 5Os +20H_{2}O$\enspace$→$\enspace$ 8Mn^{2+} +3212H_{2}O + 5OsO_{4} +40H^{+}+40e^{}
This gives the final balanced equation:8MnO_{4}^{} + 24H^{+} + 5Os + $\enspace$→$\enspace$ 8Mn^{2+} + 12H_{2}O + 5OsO_{4}

Intro Lesson
Using halfequations

1.
Balance the halfequations in the conditions stated.
Balance the halfequations below for the given conditions.