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Half-equations show only what happens to either the oxidised or the reduced species in a redox reaction. |
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Half-equations
The oxidised species in a reaction loses electrons. This may be written down as a process that doesn't show where the electrons go to, or come from. Such an equation is termed a half-equation, as it only tells half of the story.
For the reaction:
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CuSO4(aq) + Zn(s) →
ZnSO4(aq) + Cu(s)
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Electrons are transferred from the zinc metal to the copper(II) ions in solution. The process of reduction (addition of electrons) that happens to the copper (II) ions can be represented by the following half-equation:
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Cu2+(aq) + 2e →
Cu(s)
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Notice that there is no mention of the zinc in this half-equation. However, a similar half-equation could be written to represent the oxidation of Zinc in the reaction:
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Zn(s) - 2e →
Zn2+(aq)
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This could also be written as:
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Zn(s) →
Zn2+(aq) + 2e
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Notice also that the sulfate ions do not appear in either half-equation. This is because they are not involved in the original reaction. The sulfate ions are merely spectators to the whole process. At the start of the reaction they are ions in solution and they remain unchanged at the end of the reaction. For this reason they are sometimes called 'spectator ions'.
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Example: Write the half-equation for the reduction that occurs in the following reaction:
Reduction is gain of electrons. The hydrogen ions turn to hydrogen gas by gaining electrons:
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Constructing the ionic equation
The full equation for a reaction can be constructed from the two half-equations by adding them together. However, before two half-equations can be added together the number of electrons in each half-equation must be made the same.
It is important to note here that only reduction half-equations can be added to oxidation half-equations and vice versa. It would be impossible to add two reduction half-equations together as we would end up with electrons that have no home to go to.
For the following two half-equations we can construct the full (ionic) equation by simple addition. We add each left hand side together, then we add each right hand side together, as if it were a simple maths sum:
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Cu2+(aq) + 2e →
Cu(s) Cu2+(aq) + Zn(s) →
Zn2+(aq) + Cu(s)
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The electrons that appear on either side of the reaction arrow cancel out, leaving only the ions that react.
The above reaction is uncomplicated, as the number of electrons on each side of the reaction cancel out. The following example shows a situation in which the electrons do not initially cancel out, in which one half-equation has to be manipulated to equalise the electrons.
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half-equation 1: Cu(s)
→ Cu2+(aq)
+ 2e
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half-equation 2: Ag+(aq)
+ 1e → Ag(s)
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It should be seen that if we simply add these equations together there will be one extra electron remaining on the right hand side. Before the equations can be added, the second equation must be multiplied through by 2 to give:
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half-equation 2: 2Ag+(aq)
+ 2e → 2Ag(s)
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The half-equations can now be added to give:
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2Ag+(aq) + Cu(s) →
2Ag(s) + Cu2+(aq)
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This is now the balanced ionic redox reaction.
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Example: Given the two following half-equations, construct the full ionic equation for the redox reaction:
We can see that the number of electrons is the same on both sides so the half-equations can be added together directly:
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Example: Given the two following half-equations, construct the full ionic equation for the redox reaction:
The number of electrons is NOT the same on both sides so the half-equations can NOT be added together directly. First the silver half-equation must be doubled to make the electrons the same in both equations:
Now they can be added together and the electrons cancelled out:
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