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

reduction (balanced): MnO 4 ( a q ) + 4H + ( a q ) + 3e MnO 2 ( s ) + 2H 2 O ( l )

Make sure to check the half-reaction:

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

Collecting what we have so far:

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

In this case, both half reactions involve the same number of electrons; therefore, simply add the two half-reactions together.

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

Checking each side of the equation:

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

This is the balanced equation in acidic solution. For a basic solution, add one hydroxide ion to each side and simplify:

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

Checking each side of the equation:

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

This is the balanced equation in basic solution.

Check your learning

Balance the following in the type of solution indicated.

(a) H 2 + Cu 2+ Cu (acidic solution)

(b) H 2 + Cu(OH) 2 Cu (basic solution)

(c) Fe + Ag + Fe 2+ + Ag

(d) Identify the oxidizing agents in reactions (a), (b), and (c).

(e) Identify the reducing agents in reactions (a), (b), and (c).

Answer:

(a) H 2 ( g ) + Cu 2+ ( a q ) 2 H + ( a q ) + Cu ( s ) ; (b) H 2 ( g ) + Cu(OH) 2 ( s ) 2H 2 O ( l ) + Cu ( s ) ; (c) Fe ( s ) + 2Ag + ( a q ) Fe 2+ ( a q ) + 2Ag ( s ) ; (d) oxidizing agent = species reduced: Cu 2+ , Cu(OH) 2 , Ag + (e) reducing agent = species oxidized: H 2 , H 2 , Fe.

Key concepts and summary

An electric current consists of moving charge. The charge may be in the form of electrons or ions. Current flows through an unbroken or closed circular path called a circuit. The current flows through a conducting medium as a result of a difference in electrical potential between two points in a circuit. Electrical potential has the units of energy per charge. In SI units, charge is measured in coulombs (C), current in amperes ( A = C s ) , and electrical potential in volts ( V = J C ) .

Oxidation is the loss of electrons, and the species that is oxidized is also called the reducing agent. Reduction is the gain of electrons, and the species that is reduced is also called the oxidizing agent. Oxidation-reduction reactions can be balanced using the half-reaction method. In this method, the oxidation-reduction reaction is split into an oxidation half-reaction and a reduction half-reaction. The oxidation half-reaction and reduction half-reaction are then balanced separately. Each of the half-reactions must have the same number of each type of atom on both sides of the equation and show the same total charge on each side of the equation. Charge is balanced in oxidation half-reactions by adding electrons as products; in reduction half-reactions, charge is balanced by adding electrons as reactants. The total number of electrons gained by reduction must exactly equal the number of electrons lost by oxidation when combining the two half-reactions to give the overall balanced equation. Balancing oxidation-reduction reaction equations in aqueous solutions frequently requires that oxygen or hydrogen be added or removed from a reactant. In acidic solution, hydrogen is added by adding hydrogen ion (H + ) and removed by producing hydrogen ion; oxygen is removed by adding hydrogen ion and producing water, and added by adding water and producing hydrogen ion. A balanced equation in basic solution can be obtained by first balancing the equation in acidic solution, and then adding hydroxide ion to each side of the balanced equation in such numbers that all the hydrogen ions are converted to water.

Chemistry end of chapter exercises

If a 2.5 A current is run through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?

5.3 × 10 3 C

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For the scenario in the previous question, how many electrons moved through the circuit?

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For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring.

(a) Fe 3+ + 3e Fe

(b) Cr Cr 3+ + 3e

(c) MnO 4 2− MnO 4 + e

(d) Li + + e Li

(a) reduction; (b) oxidation; (c) oxidation; (d) reduction

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For each of the following unbalanced half-reactions, determine whether an oxidation or reduction is occurring.

(a) Cl Cl 2

(b) Mn 2+ MnO 2

(c) H 2 H +

(d) NO 3 NO

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Given the following pairs of balanced half-reactions, determine the balanced reaction for each pair of half-reactions in an acidic solution.

(a) Ca Ca 2+ + 2e , F 2 + 2e 2F

(b) Li Li + + e , Cl 2 + 2e 2Cl

(c) Fe Fe 3+ + 3e , Br 2 + 2e 2Br

(d) Ag Ag + + e , MnO 4 + 4H + + 3e MnO 2 + 2H 2 O

(a) F 2 + Ca 2F + Ca 2+ ; (b) Cl 2 + 2Li 2Li + + 2Cl ; (c) 3Br 2 + 2Fe 2Fe 3+ + 6Br ; (d) MnO 4 + 4H + + 3Ag 3Ag + + MnO 2 + 2H 2 O

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Balance the following in acidic solution:

(a) H 2 O 2 + Sn 2+ H 2 O + Sn 4+

(b) PbO 2 + Hg Hg 2 2+ + Pb 2+

(c) Al + Cr 2 O 7 2− Al 3+ + Cr 3+

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Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.

Oxidized: (a) Sn 2+ ; (b) Hg; (c) Al; reduced: (a) H 2 O 2 ; (b) PbO 2 ; (c) Cr 2 O 7 2− ; oxidizing agent: (a) H 2 O 2 ; (b) PbO 2 ; (c) Cr 2 O 7 2− ; reducing agent: (a) Sn 2+ ; (b) Hg; (c) Al

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Balance the following in basic solution:

(a) SO 3 2− ( a q ) + Cu(OH) 2 ( s ) SO 4 2− ( a q ) + Cu(OH) ( s )

(b) O 2 ( g ) + Mn(OH) 2 ( s ) MnO 2 ( s )

(c) NO 3 ( a q ) + H 2 ( g ) NO ( g )

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

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Identify the species that was oxidized, the species that was reduced, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.

Oxidized = reducing agent: (a) SO 3 2− ; (b) Mn(OH) 2 ; (c) H 2 ; (d) Al; reduced = oxidizing agent: (a) Cu(OH) 2 ; (b) O 2 ; (c) NO 3 ; (d) CrO 4 2−

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Why is it not possible for hydroxide ion (OH ) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in acidic solution?

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Why is it not possible for hydrogen ion (H + ) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in basic solution?

In basic solution, [OH ]>1 × 10 −7 M >[H + ]. Hydrogen ion cannot appear as a reactant because its concentration is essentially zero. If it were produced, it would instantly react with the excess hydroxide ion to produce water. Thus, hydrogen ion should not appear as a reactant or product in basic solution.

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Why must the charge balance in oxidation-reduction reactions?

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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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