Hi guys! I am starting #3 and I understand how to look at the individual oxidation states of each element in the reaction, but I am struggling with how to translate that into balancing the equation. If anyone can help guide me how to work through this type of problem, that would be amazing - thank you!
"NO is oxidized to NO−3 and Ag+ is reduced to Ag. Balance the following reaction: NO + Ag(+) -> NO3(-) + Ag"
Week 7/8 #3
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Re: Week 7/8 #3
Hi! First we can split the redox reaction into two half reactions.
Ag+ + 1e- → Ag since Ag+ gains one e- and is reduced to Ag.
NO → NO3- + 3e- since the oxidation state of N in NO is +2 and +5 in NO3-, so NO loses 3e- and is oxidized to NO3-.
The Ag reaction is already balanced so we need to balance the NO equation. I'm not sure what your question says, but mine said that this reaction occurs in a basic solution. Since it occurs in a basic solution, you can add H2O and OH- to the equation to balance the O. However, it is usually easier to treat it like an acidic solution and then convert it to a basic solution. So, in an acidic solution, you add H2O to the side that needs oxygen and H+ to the other side to balance out the H.
So NO + H2O → NO3- + H+
NO + 2H2O → NO3- + 4H+ (balances the reaction)
NO + 2H2O + 4OH- → NO3- + 4H+ + 4OH- (add however many OH- it takes to neutralize the H+)
NO + 4OH- + 2H2O → NO3- + 4H2O (the 4H+ and 4OH- form 4 H2O)
NO + 4OH- → NO3- + 2H2O (since H2O appears on both sides, you can cancel out the excess H2O)
now we have our balanced half reactions!
since the NO → NO3- oxidation reaction involves 3e- while the Ag+ → Ag reduction reaction only involves 1e-, we have to multiply the Ag+ → Ag reaction by 3 to get the correct number of electrons being transferred (3Ag+ → 3Ag)
now we have our two half reactions so we can add them together
NO + 3Ag+ + 4OH- → NO3- + 3Ag + 2H2O
Hope this helps! :)
Ag+ + 1e- → Ag since Ag+ gains one e- and is reduced to Ag.
NO → NO3- + 3e- since the oxidation state of N in NO is +2 and +5 in NO3-, so NO loses 3e- and is oxidized to NO3-.
The Ag reaction is already balanced so we need to balance the NO equation. I'm not sure what your question says, but mine said that this reaction occurs in a basic solution. Since it occurs in a basic solution, you can add H2O and OH- to the equation to balance the O. However, it is usually easier to treat it like an acidic solution and then convert it to a basic solution. So, in an acidic solution, you add H2O to the side that needs oxygen and H+ to the other side to balance out the H.
So NO + H2O → NO3- + H+
NO + 2H2O → NO3- + 4H+ (balances the reaction)
NO + 2H2O + 4OH- → NO3- + 4H+ + 4OH- (add however many OH- it takes to neutralize the H+)
NO + 4OH- + 2H2O → NO3- + 4H2O (the 4H+ and 4OH- form 4 H2O)
NO + 4OH- → NO3- + 2H2O (since H2O appears on both sides, you can cancel out the excess H2O)
now we have our balanced half reactions!
since the NO → NO3- oxidation reaction involves 3e- while the Ag+ → Ag reduction reaction only involves 1e-, we have to multiply the Ag+ → Ag reaction by 3 to get the correct number of electrons being transferred (3Ag+ → 3Ag)
now we have our two half reactions so we can add them together
NO + 3Ag+ + 4OH- → NO3- + 3Ag + 2H2O
Hope this helps! :)
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Re: Week 7/8 #3
Kayla Law 2D wrote:Hi! First we can split the redox reaction into two half reactions.
Ag+ + 1e- → Ag since Ag+ gains one e- and is reduced to Ag.
NO → NO3- + 3e- since the oxidation state of N in NO is +2 and +5 in NO3-, so NO loses 3e- and is oxidized to NO3-.
The Ag reaction is already balanced so we need to balance the NO equation. I'm not sure what your question says, but mine said that this reaction occurs in a basic solution. Since it occurs in a basic solution, you can add H2O and OH- to the equation to balance the O. However, it is usually easier to treat it like an acidic solution and then convert it to a basic solution. So, in an acidic solution, you add H2O to the side that needs oxygen and H+ to the other side to balance out the H.
So NO + H2O → NO3- + H+
NO + 2H2O → NO3- + 4H+ (balances the reaction)
NO + 2H2O + 4OH- → NO3- + 4H+ + 4OH- (add however many OH- it takes to neutralize the H+)
NO + 4OH- + 2H2O → NO3- + 4H2O (the 4H+ and 4OH- form 4 H2O)
NO + 4OH- → NO3- + 2H2O (since H2O appears on both sides, you can cancel out the excess H2O)
now we have our balanced half reactions!
since the NO → NO3- oxidation reaction involves 3e- while the Ag+ → Ag reduction reaction only involves 1e-, we have to multiply the Ag+ → Ag reaction by 3 to get the correct number of electrons being transferred (3Ag+ → 3Ag)
now we have our two half reactions so we can add them together
NO + 3Ag+ + 4OH- → NO3- + 3Ag + 2H2O
Hope this helps! :)
YES!!! This helped me SO much, I can't even explain! Thank you for walking through everything and for taking the time to answer in such depth! I really appreciate it :)
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Re: Week 7/8 #3
Thank you for these instructions! Very helpful. I am still a bit confused about how we know that NO loses three electrons.
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Re: Week 7/8 #3
I was having trouble with this problem and your walkthrough was super helpful! Thank you so much.
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Re: Week 7/8 #3
This walkthrough was super helpful! Thank you.
I believe we just start by determining the oxidation numbers (NO and NO3- here), then you compare them to see how they've changed. If I understand it right: positive oxidation numbers = negative/loss of electrons, and negative oxidation numbers = positive/gain in electrons.
In this case, NO has +2 and NO3- has +5 so there is a difference of 3, and since it'd be 2+3=5 that's 3 electrons lost.
The pdf below walks through it and has some practice problems & answers at the bottom if you want to take a look at that too. Hope this helps some.
https://www.mtsu.edu/chemistry/chem1010 ... uction.pdf
arisawaters2D wrote:Thank you for these instructions! Very helpful. I am still a bit confused about how we know that NO loses three electrons.
I believe we just start by determining the oxidation numbers (NO and NO3- here), then you compare them to see how they've changed. If I understand it right: positive oxidation numbers = negative/loss of electrons, and negative oxidation numbers = positive/gain in electrons.
In this case, NO has +2 and NO3- has +5 so there is a difference of 3, and since it'd be 2+3=5 that's 3 electrons lost.
The pdf below walks through it and has some practice problems & answers at the bottom if you want to take a look at that too. Hope this helps some.
https://www.mtsu.edu/chemistry/chem1010 ... uction.pdf
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Re: Week 7/8 #3
Thank you so much for this post, I was so confused but the in-depth explanation was so helpful!!!
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Re: Week 7/8 #3
Kayla Law 2D wrote:Hi! First we can split the redox reaction into two half reactions.
Ag+ + 1e- → Ag since Ag+ gains one e- and is reduced to Ag.
NO → NO3- + 3e- since the oxidation state of N in NO is +2 and +5 in NO3-, so NO loses 3e- and is oxidized to NO3-.
The Ag reaction is already balanced so we need to balance the NO equation. I'm not sure what your question says, but mine said that this reaction occurs in a basic solution. Since it occurs in a basic solution, you can add H2O and OH- to the equation to balance the O. However, it is usually easier to treat it like an acidic solution and then convert it to a basic solution. So, in an acidic solution, you add H2O to the side that needs oxygen and H+ to the other side to balance out the H.
So NO + H2O → NO3- + H+
NO + 2H2O → NO3- + 4H+ (balances the reaction)
NO + 2H2O + 4OH- → NO3- + 4H+ + 4OH- (add however many OH- it takes to neutralize the H+)
NO + 4OH- + 2H2O → NO3- + 4H2O (the 4H+ and 4OH- form 4 H2O)
NO + 4OH- → NO3- + 2H2O (since H2O appears on both sides, you can cancel out the excess H2O)
now we have our balanced half reactions!
since the NO → NO3- oxidation reaction involves 3e- while the Ag+ → Ag reduction reaction only involves 1e-, we have to multiply the Ag+ → Ag reaction by 3 to get the correct number of electrons being transferred (3Ag+ → 3Ag)
now we have our two half reactions so we can add them together
NO + 3Ag+ + 4OH- → NO3- + 3Ag + 2H2O
Hope this helps! :)
Thank you so much for your explanation! It is super helpful!!!
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