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oxidation (balanced): 2Cr 3+ ( a q ) + 7H 2 O ( l ) Cr 2 O 7 2− ( a q ) + 14H + ( a q ) + 6e

Checking the half-reaction:

Cr: Does ( 2 × 1 ) = ( 1 × 2 ) ? Yes . H: Does ( 7 × 2 ) = ( 14 × 1 ) ? Yes . O: Does ( 7 × 1) = ( 1 × 7 ) ? Yes . Charge: Does [ 2 × ( +3 ) ] = [ 1 × ( −2 ) + 14 × ( +1 ) + 6 × ( −1 ) ] ? Yes .

Now work on the reduction. It is necessary to convert the four oxygen atoms in the permanganate into four water molecules. To do this, add eight H + to convert the oxygen into four water molecules:

reduction (unbalanced): MnO 4 ( a q ) + 8H + ( a q ) Mn 2+ ( a q ) + 4H 2 O ( l )

Then add five electrons to the left side to balance the charge:

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

Make sure to check the half-reaction:

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 .

Collecting what we have so far:

oxidation: 2 Cr 3+ ( a q ) + 7 H 2 O ( l ) Cr 2 O 7 2− ( a q ) + 14 H + ( a q ) + 6 e reduction: MnO 4 ( a q ) + 8 H + ( a q ) + 5 e Mn 2+ ( a q ) + 4 H 2 O ( l )

The least common multiple for the electrons is 30, so multiply the oxidation half-reaction by five, the reduction half-reaction by six, combine, and simplify:

10 Cr 3+ ( a q ) + 35 H 2 O ( l ) + 6 MnO 4 ( a q ) + 48 H + ( a q ) 5 Cr 2 O 7 2− ( a q ) + 70 H + ( a q ) + 6 Mn 2+ ( a q ) + 24 H 2 O ( l )
10 Cr 3+ ( a q ) + 11 H 2 O ( l ) + 6 MnO 4 ( a q ) 5 Cr 2 O 7 2− ( a q ) + 22 H + ( a q ) + 6 Mn 2+ ( a q )

Checking each side of the equation:

Mn: Does ( 6 × 1 ) = ( 6 × 1 ) ? Yes . Cr: Does ( 10 × 1 ) = ( 5 × 2 ) ? Yes . H: Does ( 11 × 2 ) = ( 22 × 1 ) ? Yes . O: Does ( 11 × 1 + 6 × 4 ) = ( 5 × 7 ) ? Yes . Charge: Does [ 10 × ( +3 ) + 6 × ( −1 ) ] = [ 5 × ( −2 ) + 22 × ( +1 ) + 6 × ( +2 ) ] ? Yes .

This is the balanced equation in acidic solution.

Check your learning

Balance the following equation in acidic solution:

Hg 2 2+ + Ag Hg + Ag +

Answer:

Hg 2 2+ ( a q ) + 2Ag ( s ) 2Hg ( l ) + 2Ag + ( a q )

Balancing basic oxidation-reduction reactions

Balance the following reaction equation in basic solution:

MnO 4 ( a q ) + Cr ( OH ) 3 ( s ) MnO 2 ( s ) + CrO 4 2− ( a q )

Solution

This is an oxidation-reduction reaction, so start by collecting the species given into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction

oxidation (unbalanced): Cr(OH) 3 ( s ) CrO 4 2− ( a q ) reduction (unbalanced): MnO 4 ( a q ) MnO 2 ( s )

Starting with the oxidation half-reaction, we can balance the chromium

oxidation (unbalanced): Cr(OH) 3 ( s ) CrO 4 2− ( a q )

In acidic solution, we can use or generate hydrogen ions (H + ). Adding one water molecule to the left side provides the necessary oxygen; the “left over” hydrogen appears as five H + on the right side:

oxidation (unbalanced): Cr(OH) 3 ( s ) + H 2 O ( l ) CrO 4 2− ( a q ) + 5H + ( a q )

The left side of the equation has a total charge of [0], and the right side a total charge of [−2 + 5 × (+1) = +3]. The difference is three, adding three electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):

oxidation (balanced): Cr(OH) 3 ( s ) + H 2 O ( l ) CrO 4 2− ( a q ) + 5H + ( a q ) + 3e

Checking the half-reaction:

Cr: Does ( 1 × 1 ) = ( 1 × 1 ) ? Yes . H: Does ( 1 × 3 + 1 × 2 ) = ( 5 × 1 ) ? Yes . O: Does ( 1 × 3 + 1 × 1 ) = ( 4 × 1 ) ? Yes . Charge: Does [ 0 = [ 1 × ( −2 ) + 5 × ( +1 ) + 3 × ( −1 ) ] ? Yes .

Now work on the reduction. It is necessary to convert the four O atoms in the MnO 4 minus the two O atoms in MnO 2 into two water molecules. To do this, add four H + to convert the oxygen into two water molecules:

reduction (unbalanced): MnO 4 ( a q ) + 4H + ( a q ) MnO 2 ( s ) + 2H 2 O ( l )
Practice Key Terms 6

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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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