How to balance a complex chemical equation?
I've always struggled in terms of knowing where to start when balancing long equations, eg:
As2O3 + Zn +HNO3 --> ASH3 + ZN(NO3)2 + H20
I always tend to leave singular elements last because theyre easiest to balance (eg Zinc in this case) but where do I start for the others?
As2O3 + Zn +HNO3 --> ASH3 + ZN(NO3)2 + H20
I always tend to leave singular elements last because theyre easiest to balance (eg Zinc in this case) but where do I start for the others?
Reply 1
Original post by usernihilnomen
I've always struggled in terms of knowing where to start when balancing long equations, eg:
As2O3 + Zn +HNO3 --> ASH3 + ZN(NO3)2 + H20
I always tend to leave singular elements last because theyre easiest to balance (eg Zinc in this case) but where do I start for the others?
As2O3 + Zn +HNO3 --> ASH3 + ZN(NO3)2 + H20
I always tend to leave singular elements last because theyre easiest to balance (eg Zinc in this case) but where do I start for the others?
Stepwise approach is the most effective method to balance equation redox. Let us work through your example together:
Unbalanced Equation:
As2O3 + Zn + HNO3 ==> AsH3 + Zn(NO3)2 + H2O
a) Identify Redox Pairs
In this reaction:
As is reduced (from +3 in As2O3 to -3 in AsH3)
Zn is oxidized (from 0 to +2 in Zn(NO3)2)
b) Oxidation Number
For example in As2O3: +3 (since O is -2)
Zn: 0 (as a pure element)
In AsH3: -3 (with H being +1)
Zn in Zn(NO3)2: +2
c) : Write the Half-Reactions
For Reduction (As):
As2O3 ==> AsH3
Balance As: As2O3 ==> 2AsH3
H2O with O: As2O3 ==> 2AsH3 + 3H2O
– If it’s an acidic solution, balance H with H^+:
As2O3+ 12H^+ ==> 2AsH3 + 3H2O
− Balance charge: As2O3 + 12H^+ + 12e⁻ ==> 2AsH3 + 3H2O
For Oxidation (Zn):
Zn ==> Zn^2+
Zn==> Zn^2+ + 2e⁻
d) : Balance Electrons
Multiply the oxidation reaction by 6 in order to have 12 electrons, the same amount as in the reduction step:
6Zn ==> 6Zn^2+ + 12e⁻
Now we can combine the half-reactions:
2AsH3+ 3H2O + 6Zn^2+ + 12e⁻ + As2O3 + 12H^+
Cancel out the electrons:
As2O3 + 12H^+ + 6Zn → 2AsH3 + 3H2O + 6Zn^2+
e) : Introduce NO3^- (from HNO3)
The 6Zn^2+ will react with 12NO3^- to produce a total of 6 Zn(NO3)2:
6Zn^2+ + 12NO3^- ==> 6Zn(NO3)2
Therefore we need 12 HNO3 to provide the H^+ and NO3^-:
As2O3+ 6Zn + 12HNO3 ==> 2AsH3 + 6Zn(NO3)2 + 3H2O
f) : Check the Balance
As: 2 on both sides
Zn: 6 on both sides
− H: 12 on the left (from HNO3) and 12 on the right (6 in AsH3 and 6 in H2O)
N: 12 in Zn(NO3)2 on both sides
– O: 39 in total both sides (3 from As2O3 +36 from HNO3 =39; 36 in Zn(NO3)2 and +3 in H2O =39)
Final Balanced Equation:
As2O3 + 6Zn + 12HNO3 ==> 2AsH3 + 6Zn(NO3)2 + 3H2O
Key Points:
Key Points:
1.
Figure out the redox couples and write half reactions.
2.
Balance atoms (other than H & O) before; then H2O and then H^+ for O and H.
3.
Then balance charge with electrons.
4.
Combine the half-reactions and cancel species.
5.
Add spectator ions (such as NO3^-) at the end. From the web: Spectator ions are ions that appear on both sides of a chemical equation and do not participate in the reaction. They are also known as "spectator" ions because they "watch" the other ions react.
Sandro
(edited 1 month ago)
Reply 2
Original post by usernihilnomen
I've always struggled in terms of knowing where to start when balancing long equations, eg:
As2O3 + Zn +HNO3 --> ASH3 + ZN(NO3)2 + H20
I always tend to leave singular elements last because theyre easiest to balance (eg Zinc in this case) but where do I start for the others?
As2O3 + Zn +HNO3 --> ASH3 + ZN(NO3)2 + H20
I always tend to leave singular elements last because theyre easiest to balance (eg Zinc in this case) but where do I start for the others?
I’d consider the oxidation states of the important elements first.
Arsenic is +3 in As2O3 and -3 in AsH3. The oxidation state of arsenic changes by (+3) - (-3) = 6, so each arsenic must gain 6 electrons.
Zinc is 0 in Zn and +2 in Zn(NO3)2. The oxidation state therefore changes by 2 and so each zinc loses two electrons.
Nitrogen, oxygen and hydrogen maintain their oxidation states throughout and so are not important to consider for the time being.
We can deduce that the ratio of As2O3 to Zn reacting must be 1:6, since there are two arsenics per As2O3 and so 12 electrons must be gained. You need 6 zinc atoms to release this number of electrons.
Since each zinc ends up binding to 2 nitrate ions in the product, we can deduce that there must be 6 x 2 = 12 HNO3’s on the LHS:
As2O3 + 6Zn + 12HNO3 —> …
We can see that there are 2 arsenics on the LHS (so there must be 2 AsH3’s on the RHS) and 6 zincs on the LHS (so there must be 6 Zn(NO3)2’s on the RHS). We therefore have the following equation (which isn’t yet balanced):
As2O3 + 6Zn + 12HNO3 —> 2AsH3 + 6Zn(NO3)2
We can see that on the LHS there are 12 hydrogens and 39 oxygens, but only 6 hydrogens and 36 oxygens on the RHS. This means adding 6 more hydrogens and 3 oxygens on the RHS in the form of 3 water molecules:
As2O3 + 6Zn + 12HNO3 —> 2AsH3 + 6Zn(NO3)2 + 3H2O
And this is the correct answer. This particular reaction is a fun one as it was used in the “Marsh test” to prove definitively that arsenic was used in poisonings.
Reply 3
Original post by TypicalNerd
I’d consider the oxidation states of the important elements first.
Arsenic is +3 in As2O3 and -3 in AsH3. The oxidation state of arsenic changes by (+3) - (-3) = 6, so each arsenic must gain 6 electrons.
Zinc is 0 in Zn and +2 in Zn(NO3)2. The oxidation state therefore changes by 2 and so each zinc loses two electrons.
Nitrogen, oxygen and hydrogen maintain their oxidation states throughout and so are not important to consider for the time being.
We can deduce that the ratio of As2O3 to Zn reacting must be 1:6, since there are two arsenics per As2O3 and so 12 electrons must be gained. You need 6 zinc atoms to release this number of electrons.
Since each zinc ends up binding to 2 nitrate ions in the product, we can deduce that there must be 6 x 2 = 12 HNO3’s on the LHS:
As2O3 + 6Zn + 12HNO3 —> …
We can see that there are 2 arsenics on the LHS (so there must be 2 AsH3’s on the RHS) and 6 zincs on the LHS (so there must be 6 Zn(NO3)2’s on the RHS). We therefore have the following equation (which isn’t yet balanced):
As2O3 + 6Zn + 12HNO3 —> 2AsH3 + 6Zn(NO3)2
We can see that on the LHS there are 12 hydrogens and 39 oxygens, but only 6 hydrogens and 36 oxygens on the RHS. This means adding 6 more hydrogens and 3 oxygens on the RHS in the form of 3 water molecules:
As2O3 + 6Zn + 12HNO3 —> 2AsH3 + 6Zn(NO3)2 + 3H2O
And this is the correct answer. This particular reaction is a fun one as it was used in the “Marsh test” to prove definitively that arsenic was used in poisonings.
Arsenic is +3 in As2O3 and -3 in AsH3. The oxidation state of arsenic changes by (+3) - (-3) = 6, so each arsenic must gain 6 electrons.
Zinc is 0 in Zn and +2 in Zn(NO3)2. The oxidation state therefore changes by 2 and so each zinc loses two electrons.
Nitrogen, oxygen and hydrogen maintain their oxidation states throughout and so are not important to consider for the time being.
We can deduce that the ratio of As2O3 to Zn reacting must be 1:6, since there are two arsenics per As2O3 and so 12 electrons must be gained. You need 6 zinc atoms to release this number of electrons.
Since each zinc ends up binding to 2 nitrate ions in the product, we can deduce that there must be 6 x 2 = 12 HNO3’s on the LHS:
As2O3 + 6Zn + 12HNO3 —> …
We can see that there are 2 arsenics on the LHS (so there must be 2 AsH3’s on the RHS) and 6 zincs on the LHS (so there must be 6 Zn(NO3)2’s on the RHS). We therefore have the following equation (which isn’t yet balanced):
As2O3 + 6Zn + 12HNO3 —> 2AsH3 + 6Zn(NO3)2
We can see that on the LHS there are 12 hydrogens and 39 oxygens, but only 6 hydrogens and 36 oxygens on the RHS. This means adding 6 more hydrogens and 3 oxygens on the RHS in the form of 3 water molecules:
As2O3 + 6Zn + 12HNO3 —> 2AsH3 + 6Zn(NO3)2 + 3H2O
And this is the correct answer. This particular reaction is a fun one as it was used in the “Marsh test” to prove definitively that arsenic was used in poisonings.
I sincerely apologise if this is extremely obvious, but just to be clear when determining which molecules are actually reacting, it's always the ones where the oxidation states changes?
Otherwise thank you so much for your comprehensive answer!
Reply 4
Original post by Nitrotoluene
Stepwise approach is the most effective method to balance equation redox. Let us work through your example together:
Unbalanced Equation:
As2O3 + Zn + HNO3 ==> AsH3 + Zn(NO3)2 + H2O
a) Identify Redox Pairs
In this reaction:
As is reduced (from +3 in As2O3 to -3 in AsH3)
Zn is oxidized (from 0 to +2 in Zn(NO3)2)
b) Oxidation Number
For example in As2O3: +3 (since O is -2)
Zn: 0 (as a pure element)
In AsH3: -3 (with H being +1)
Zn in Zn(NO3)2: +2
c) : Write the Half-Reactions
For Reduction (As):
As2O3 ==> AsH3
Balance As: As2O3 ==> 2AsH3
H2O with O: As2O3 ==> 2AsH3 + 3H2O
– If it’s an acidic solution, balance H with H^+:
As2O3+ 12H^+ ==> 2AsH3 + 3H2O
− Balance charge: As2O3 + 12H^+ + 12e⁻ ==> 2AsH3 + 3H2O
For Oxidation (Zn):
Zn ==> Zn^2+
Zn==> Zn^2+ + 2e⁻
d) : Balance Electrons
Multiply the oxidation reaction by 6 in order to have 12 electrons, the same amount as in the reduction step:
6Zn ==> 6Zn^2+ + 12e⁻
Now we can combine the half-reactions:
2AsH3+ 3H2O + 6Zn^2+ + 12e⁻ + As2O3 + 12H^+
Cancel out the electrons:
As2O3 + 12H^+ + 6Zn → 2AsH3 + 3H2O + 6Zn^2+
e) : Introduce NO3^- (from HNO3)
The 6Zn^2+ will react with 12NO3^- to produce a total of 6 Zn(NO3)2:
6Zn^2+ + 12NO3^- ==> 6Zn(NO3)2
Therefore we need 12 HNO3 to provide the H^+ and NO3^-:
As2O3+ 6Zn + 12HNO3 ==> 2AsH3 + 6Zn(NO3)2 + 3H2O
f) : Check the Balance
As: 2 on both sides
Zn: 6 on both sides
− H: 12 on the left (from HNO3) and 12 on the right (6 in AsH3 and 6 in H2O)
N: 12 in Zn(NO3)2 on both sides
– O: 39 in total both sides (3 from As2O3 +36 from HNO3 =39; 36 in Zn(NO3)2 and +3 in H2O =39)
Final Balanced Equation:
Bye,
Sandro
Unbalanced Equation:
As2O3 + Zn + HNO3 ==> AsH3 + Zn(NO3)2 + H2O
a) Identify Redox Pairs
In this reaction:
As is reduced (from +3 in As2O3 to -3 in AsH3)
Zn is oxidized (from 0 to +2 in Zn(NO3)2)
b) Oxidation Number
For example in As2O3: +3 (since O is -2)
Zn: 0 (as a pure element)
In AsH3: -3 (with H being +1)
Zn in Zn(NO3)2: +2
c) : Write the Half-Reactions
For Reduction (As):
As2O3 ==> AsH3
Balance As: As2O3 ==> 2AsH3
H2O with O: As2O3 ==> 2AsH3 + 3H2O
– If it’s an acidic solution, balance H with H^+:
As2O3+ 12H^+ ==> 2AsH3 + 3H2O
− Balance charge: As2O3 + 12H^+ + 12e⁻ ==> 2AsH3 + 3H2O
For Oxidation (Zn):
Zn ==> Zn^2+
Zn==> Zn^2+ + 2e⁻
d) : Balance Electrons
Multiply the oxidation reaction by 6 in order to have 12 electrons, the same amount as in the reduction step:
6Zn ==> 6Zn^2+ + 12e⁻
Now we can combine the half-reactions:
2AsH3+ 3H2O + 6Zn^2+ + 12e⁻ + As2O3 + 12H^+
Cancel out the electrons:
As2O3 + 12H^+ + 6Zn → 2AsH3 + 3H2O + 6Zn^2+
e) : Introduce NO3^- (from HNO3)
The 6Zn^2+ will react with 12NO3^- to produce a total of 6 Zn(NO3)2:
6Zn^2+ + 12NO3^- ==> 6Zn(NO3)2
Therefore we need 12 HNO3 to provide the H^+ and NO3^-:
As2O3+ 6Zn + 12HNO3 ==> 2AsH3 + 6Zn(NO3)2 + 3H2O
f) : Check the Balance
As: 2 on both sides
Zn: 6 on both sides
− H: 12 on the left (from HNO3) and 12 on the right (6 in AsH3 and 6 in H2O)
N: 12 in Zn(NO3)2 on both sides
– O: 39 in total both sides (3 from As2O3 +36 from HNO3 =39; 36 in Zn(NO3)2 and +3 in H2O =39)
Final Balanced Equation:
As2O3 + 6Zn + 12HNO3 ==> 2AsH3 + 6Zn(NO3)2 + 3H2O
Key Points:
Key Points:
1.
Figure out the redox couples and write half reactions.
2.
Balance atoms (other than H & O) before; then H2O and then H^+ for O and H.
3.
Then balance charge with electrons.
4.
Combine the half-reactions and cancel species.
5.
Add spectator ions (such as NO3^-) at the end. From the web: Spectator ions are ions that appear on both sides of a chemical equation and do not participate in the reaction. They are also known as "spectator" ions because they "watch" the other ions react.
Sandro
Thank you so much!
Reply 5
Original post by usernihilnomen
I sincerely apologise if this is extremely obvious, but just to be clear when determining which molecules are actually reacting, it's always the ones where the oxidation states changes?
Otherwise thank you so much for your comprehensive answer!
Otherwise thank you so much for your comprehensive answer!
The ones where the oxidation states change are the easiest starting point. It’s not that anything that doesn’t change oxidation states is not a reactant - admittedly my wording in my first post could have been better.
Because you can easily tell how many electrons the reducing agent loses and how many the oxidising agent gain from the respective changes in oxidation state, it’s quite easy to deduce the reacting ratio of oxidising agent to reducing agent. The rest should then follow easily from there.
Reply 6
Original post by TypicalNerd
The ones where the oxidation states change are the easiest starting point. It’s not that anything that doesn’t change oxidation states is not a reactant - admittedly my wording in my first post could have been better.
Because you can easily tell how many electrons the reducing agent loses and how many the oxidising agent gain from the respective changes in oxidation state, it’s quite easy to deduce the reacting ratio of oxidising agent to reducing agent. The rest should then follow easily from there.
Because you can easily tell how many electrons the reducing agent loses and how many the oxidising agent gain from the respective changes in oxidation state, it’s quite easy to deduce the reacting ratio of oxidising agent to reducing agent. The rest should then follow easily from there.
I see, thank you!
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