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Electrochemistry II: Balancing Equations
This chapter concerns the application of these four principles to the balancing of equations for electron-transfer reactions.
BALANCING EQUATIONS, WITH HALF-REACTIONS GIVEN
The simplest situation that exists for balancing electron-transfer equations is
the one in which a table of standard electrode potentials is at hand, and the two
needed half-reactions are included in it. The following problem illustrates this
situation.
PROBLEM:
Write a balanced ionic equation for the reaction between MnOi" and H 2 S in acid
solution.
SOLUTION:
The reaction is spontaneous because, in Table 17-1, the reducing half-reaction
2e~ + 2H
+ + S <=t H 2 S
lies above the oxidizing half-reaction
5e- + 8H
+ + MnO 4 - «=* 4H 2 O + Mn
2+
In order to combine these two half-reactions to give the complete reaction, we
must multiply each one by a factor that will yield the same number of electrons
lost as gained. The factor 5 is needed for the H 2 S half-reaction, and the factor 2 is
needed for the MnO«T half-reaction, in order to provide a loss of 10 electrons by
H 2 S and a gain of 10 by MnO 4 ~:
10e- + 16 H
+ + 2MnO 4 - ?± 8H 2 O + 2Mn
2+
10e- + 10 H
+ + 5S <=i 5H 2 S
Now if we subtract the second (reducing) half-reaction from the first (oxidizing)
half-reaction, the 10 electrons cancel to give
5H 2 S + 2MnO 4 - + 16 H
+ -» 5S + 2Mn
2+ + 10H
+ + 8H 2 O
This equation can be simplified by subtracting 10 H
+ from each side to give
5H 2 S + 2MnO 4 - + 6H+ -> 5S + 2Mn
2+ + 8H 2 O
We can make the following general statement. To balance any electrontransfer equation, you must subtract the reducing half-reaction equation from the
oxidizing half-reaction equation after the two equations have first been written to
show the same number of electrons. Simplify the final equation, if needed.