chapter 15 electro- chemistry 15.3 balancing redox equations

CHAPTER 15

Electro-chemistry

15.3 Balancing Redox Equations

2 15.3 Balancing Redox Equations

Redox reactions

Zn(s)

CuSO4(aq)

Cu(s) deposit

Zn(s) + CuSO4 → ZnSO4(aq) + Cu(s)


Redox reactions


We saw that this is a redox reaction in which:some elements lose electrons; they are oxidized

other elements gain electrons; they are reduced

We learned how to determine oxidation numbers


Redox reactions


We saw that this is a redox reaction in which:some elements lose electrons; they are oxidized

other elements gain electrons; they are reduced

We learned how to determine oxidation numbers

Now we look at how to balance redox reactions


Two methods

At the end, both mass and charge have to be balanced

There are two methods:1) The oxidation number method

2) The half-reaction method


Step 1 Assign oxidation numbers for all atoms

Step 2 Identify the atoms that are oxidized, and atoms that are reduced

Step 3 Adjust coefficients for atoms whose oxidation numbers change,

then make sure that the rule above is observed

Step 4 Check the overall mass balance

The oxidation number method

Increase in oxidation number for oxidized atoms

Decrease in oxidation number for reduced atoms

=


Using the oxidation number method, balance the equation:

HNO3(aq) + Cu2O(s) → Cu(NO3)2(aq) + NO(g) + H2O(l)




Step 1 Assign oxidation numbers for all atoms

H N O3 + Cu2 O → Cu (N O3)2 + N O + H2 O

+1 +5 –2 +1 –2 +2 +5 –2 +2 –2 +1 –2


oxidation

reduction




H N O3 + Cu2 O → Cu (N O3)2 + N O + H2 O

Oxidation numbers do not change

+1 +5 –2 +1 –2 +2 +5 –2 +2 –2 +1 –2

N is reduced its oxidation number goes from +5 to +2Cu is oxidized its oxidation number goes from +1 to +2




Step 3 Balance all atoms whose oxidation numbers have changed

H N O3 + Cu2 O → 2Cu (N O3)2 + N O + H2 O

oxidation

reduction

+1 +5 –2 +1 –2 +2 +5 –2 +2 –2 +1 –2

N is already balanced: 1 atom on each sideCu needs to be adjusted with a coefficient of 2

N is reduced its oxidation number goes from +5 to +2Cu is oxidized its oxidation number goes from +1 to +2




Step 3 Balance all atoms whose oxidation numbers have changedBalance the number of electrons (using oxidation numbers)

H N O3 + Cu2 O → 2Cu (N O3)2 + N O + H2 O

oxidation

reduction

N is reduced its oxidation number goes from +5 to +2; it gains 3 electronsCu is oxidized its oxidation number goes from +1 to +2; it loses 2 electrons

because of the coefficient

+1 +5 –2 +1 –2 +2 +5 –2 +2 –2 +1 –2





H N O3 + Cu2 O → 2Cu (N O3)2 + N O + H2 O

–2 e–

+ 3 e–


The number of electrons transferred must be the same

+1 +5 –2 +1 –2 +2 +5 –2 +2 –2 +1 –2





2H N O3 + 3Cu2 O → 6Cu (N O3)2 + 2N O + H2 O


The number of electrons transferred must be the same

(+ 3 e–) x 2 = +6 e–

+1 +5 –2 +1 –2 +2 +5 –2 +2

(–2 e–) x 3 = –6 e–





12HNO3 + 3Cu2O → 6Cu(NO3)2 + 2NO + 7H2O

H 2N 2Cu 6O 7

H 2N 14Cu 6O 39

Masses are not balanced yetStart balancing N





14HNO3 + 3Cu2O → 6Cu(NO3)2 + 2NO + 7H2O

H 14N 14Cu 6O 45

H 2N 14Cu 6O 39

Masses are not balanced yetNow balance O and H





14HNO3 + 3Cu2O → 6Cu(NO3)2 + 2NO + 7H2O

H 14N 14Cu 6O 45

H 14N 14Cu 6O 45

Masses are now balanced




14HNO3 + 3Cu2O → 6Cu(NO3)2 + 2NO + 7H2O

Answer:


Two methods




The redox reaction is split

into two half-reactions:

the oxidation reaction, and

the reduction reaction.


Step 1 Write the unbalanced equation showing explicitly all ions

The half-reaction method




Find spectator ions (unchanged oxidation numbers)






Step 3 Write down the two unbalanced half-reactions







Step 4 Balance mass with elements other than oxygen and hydrogen

Balance oxygen by adding H2O, then balance hydrogen with H+









Step 5 Balance the charge for both half-reactions by adding electrons










Step 6 Adjust coefficients to balance the number of electrons transferred











Step 7 Combine the two half-reactions, and add the spectator ions












Step 8 Simplify and check that both mass and charge are balanced



Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)


Zn(s) + Cu2+(aq) + SO4–2(aq) → Zn2+(aq) + SO4

–2(aq) + Cu(s)





0 +2 –2 +2 –2 0

oxidationreduction

spectator


–2(aq) + Cu(s)



Step 3 Write down the unbalanced half-reactions

0 +2 –2 +2 –2 0

oxidationreduction


–2(aq) + Cu(s)

Oxidation: Reduction:

Zn(s) → Zn2+(aq) Cu2+(aq) → Cu(s)





Zn(s) → Zn2+(aq)

mass already balanced mass already balanced


Cu2+(aq) → Cu(s)




Zn(s) → Zn2+(aq) + 2e–


Cu2+(aq) + 2e– → Cu(s)




Zn(s) → Zn2+(aq) + 2e–

already the same in both half-reactions


Cu2+(aq) + 2e– → Cu(s)




Zn(s) → Zn2+(aq) + 2e–



Cu2+(aq) + 2e– → Cu(s)



Step 8 Simplify and check that both mass and charge are balanced


Using the half-reaction method, we were able to determine

that 2 electrons were transferred

The full equation could not provide us with that information

chapter 15 electro- chemistry 15.3 balancing redox equations

Documents

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oxidizedother elements