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Reduction (balanced): MnO 4 ( a q ) + 8H + ( a q ) + 5e Mn 2+ ( a q ) + 4H 2 O ( l )

You should check this half-reaction for each atom type and for the charge, as well:

Mn: Does ( 1 × 1 ) = ( 1 × 1 ) ? Yes . H: Does ( 8 × 1) = ( 4 × 2 ) ? Yes . O: Does ( 1 × 4) = ( 4 × 1 ) ? Yes . Charge: Does [ 1 × ( −1 ) + 8 × ( +1 ) + 5 × ( −1 ) ] = [ 1 × ( +2 ) ] ? Yes .

Now that this half-reaction is balanced, it is easy to see it involves reduction because electrons were gained when MnO 4 was reduced to Mn 2+ . In all reduction half-reactions, electrons appear as reactants (on the left side). As discussed in the earlier chapter, the species that was reduced, MnO 4 in this case, is also called the oxidizing agent. We now have two balanced half-reactions.

oxidation: Fe 2+ ( a q ) Fe 3+ ( a q ) + e reduction: MnO 4 ( a q ) + 8H + ( a q ) + 5e Mn 2+ ( a q ) + 4H 2 O ( l )

It is now necessary to combine the two halves to produce a whole reaction. The key to combining the half-reactions is the electrons. The electrons lost during oxidation must go somewhere. These electrons go to cause reduction. The number of electrons transferred from the oxidation half-reaction to the reduction half-reaction must be equal. There can be no missing or excess electrons. In this example, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five; therefore, it is necessary to multiply every term in the oxidation half-reaction by five and every term in the reduction half-reaction by one. (In this case, the multiplication of the reduction half-reaction generates no change; however, this will not always be the case.) The multiplication of the two half-reactions by the appropriate factor followed by addition of the two halves gives

oxidation: 5 × ( Fe 2+ ( a q ) Fe 3+ ( a q ) + e ) reduction: MnO 4 ( a q ) + 8H + ( a q ) + 5e Mn 2+ ( a q ) + 4H 2 O ( l ) ¯ overall: 5Fe 2+ ( a q ) + MnO 4 ( a q ) + 8H + ( a q ) 5Fe 3+ ( a q ) + Mn 2+ ( a q ) + 4H 2 O ( l )

The electrons do not appear in the final answer because the oxidation electrons are the same electrons as the reduction electrons and they “cancel.” Carefully check each side of the overall equation to verify everything was combined correctly:

Fe: Does ( 5 × 1 ) = ( 5 × 1 ) ? Yes . Mn: Does ( 1 × 1 ) = ( 1 × 1 ) ? Yes . H: Does ( 8 × 1 ) = ( 4 × 2 ) ? Yes . O: Does ( 1 × 4 ) = ( 4 × 1 ) ? Yes . Charge: Does [ 5 × ( +2 ) + 1 × ( −1 ) + 8 × ( +1 ) ] = [ 5 × ( +3 ) + 1 × ( +2 ) ] ? Yes .

Everything checks, so this is the overall equation in acidic solution. If something does not check, the most common error occurs during the multiplication of the individual half-reactions.

Now suppose we wanted the solution to be basic. Recall that basic solutions have excess hydroxide ions. Some of these hydroxide ions will react with hydrogen ions to produce water. The simplest way to generate the balanced overall equation in basic solution is to start with the balanced equation in acidic solution, then “convert” it to the equation for basic solution. However, it is necessary to exercise caution when doing this, as many reactants behave differently under basic conditions and many metal ions will precipitate as the metal hydroxide. We just produced the following reaction, which we want to change to a basic reaction:

Practice Key Terms 6

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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