Assigning Oxidation States - AP Chemistry
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What is the oxidation state of gold in 
?
What is the oxidation state of gold in
Let's first consider the overall compound 
. Since it is negatively charged overall it is equal to 
. Applying our elementary rules of oxidation states we know that halogens generally have an oxidation state of 
, so this means chlorine has an oxidation state of 
. Since there are 
chlorines we must multiply 
to get the charge of all the chlorines together. This is equal to 
.
We can now solve for oxidation state of gold through creating an equation as shown below:
Simplifying this we get that 
, therefore the oxidation state of gold is equal to 
Let's first consider the overall compound
We can now solve for oxidation state of gold through creating an equation as shown below:
Simplifying this we get that
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Determine the oxidation number of each element in the compound 
Determine the oxidation number of each element in the compound
The oxidation number of each element in the compound 
is:
The oxidation number of chlorine is 
. There are 3 chlorine atoms present in the compound, so 
Because this is a neutral atom, the overall charge is 0. Therefore we can set 
Therefore, 
We can then solve for the oxidation number for 
The oxidation number of each element in the compound
The oxidation number of chlorine is
Because this is a neutral atom, the overall charge is 0. Therefore we can set
Therefore,
We can then solve for the oxidation number for
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Given the following equation, identify the reducing agent.
Given the following equation, identify the reducing agent.
Recall that a reducing agent is being oxidized and is losing electrons. Thus, when looking at the equation, look for which oxidation state is becoming more positive.
Start by assigning oxidation states to the elements.
For the reactants:
has an oxidation state of 
.
has an oxidation state of 
.
has an oxidation state of 
has an oxidation state of 
.
Now, assign the oxidation state of the products.
has an oxidation state of 
.
has an oxidation state of 
.
has an oxidation state of 
.
has an oxidation state of 
.
Only 
undergoes a loss of electrons. It must be the reducing agent.
Recall that a reducing agent is being oxidized and is losing electrons. Thus, when looking at the equation, look for which oxidation state is becoming more positive.
Start by assigning oxidation states to the elements.
For the reactants:
Now, assign the oxidation state of the products.
Only
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Which compound below has a nitrogen atom at a 
oxidation state?
Which compound below has a nitrogen atom at a
Let's start by going through each answer case-by-case applying the elementary rules of oxidation states we are given. The most applicable of these rules in this problem are the oxidation state of hydrogen in a compound is generally equal to 
, and the oxidation state of oxygen in a compound is generally equal to 
.
Let's start by looking at 
. We know this compound as a whole is neutrally charged (equal to 
overall charge), and applying our knowledge of the general oxidation state of hydrogen in a compound and knowing there are 
hydrogens we know that the 
hydrogens have a total charge of 
together, and nitrogen has an unknown oxidation state 
, so for our equation we have:
solving for 
gives 
, giving us the answer we need. Therefore the nitrogen atom in 
has an oxidation state of 
, which is the correct answer.
Let's still look at the other cases, however:
As this compound is uncharged we know it's net charge is equal to 
Using our knowledge of oxidation states of hydrogen and oxygen and counting the number of hydrogens and oxygen in this compound, we can determine that the total charge of all the hydrogens together is equal to 
, and that the total charge of the one oxygen is 
.
Just as we did above we can create an equation and solve for our unknown.
therefore 
, so the oxidation state of nitrogen in 
is 
. This means this isn't the right compound.
Next let's look at 
:
We can quickly apply our knowledge of oxidation states of oxygen knowing that the oxidation state of oxygen in compounds is generally 
. Since there are 
oxygens we must multiply 
to find out the total charge contributed by all the oxygens, which is 
.
So our equation for this becomes:
therefore 
and the oxidation state of nitrogen is 
, which isn't the answer we are looking for.
Finally the last compound is 
:
With this compound we can apply the rule we know about hydrogen's oxidation state in a compound being equal to 
. Since there are 
hydrogens, we must multiply 
which equals 
.
Since there are two nitrogens, our equation is:
and 
. Therefore the oxidation state of nitrogen is equal to 
.
Let's start by going through each answer case-by-case applying the elementary rules of oxidation states we are given. The most applicable of these rules in this problem are the oxidation state of hydrogen in a compound is generally equal to
Let's start by looking at
solving for
Let's still look at the other cases, however:
As this compound is uncharged we know it's net charge is equal to
Using our knowledge of oxidation states of hydrogen and oxygen and counting the number of hydrogens and oxygen in this compound, we can determine that the total charge of all the hydrogens together is equal to
Just as we did above we can create an equation and solve for our unknown.
therefore
Next let's look at
We can quickly apply our knowledge of oxidation states of oxygen knowing that the oxidation state of oxygen in compounds is generally
So our equation for this becomes:
therefore
Finally the last compound is
With this compound we can apply the rule we know about hydrogen's oxidation state in a compound being equal to
Since there are two nitrogens, our equation is:
and
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In the given reaction, which element(s) is/are oxidized?
In the given reaction, which element(s) is/are oxidized?
Let's start by looking at the equation and assigning oxidation numbers based off of the general rules we are given:
Let's start by looking at the reactants, namely 
. Knowing that an atom in its elemental form has an oxidation number of 
, 
has an oxidation number equal to 
.
Next let's look at 
. This is a neutral compound so know the oxidation number of the whole compound must equal 
. Going off of our general rules, the oxidation number of fluorine is equal 
. Since we have 3 fluorines we multiply the oxidation number by 
to get the cumulative oxidation number of all the fluorines together to get 
. So now we must solve for the oxidation of Cl, which is unknown. This is done through a simple algebraic equation: 
so 
. Therefore the oxidation number of Chlorine is 
.
Now we have the oxidation numbers of all the elements present in this equation that are on the reactant side. I recommend writing down their oxidation numbers next to each other, so the elements that are oxidized and reduced can quickly be determined.
Oxidation numbers of elements in reactants:
Now let's look at the product side of the equation and determine the oxidation numbers of the elements there. First let's look at 
. Knowing the general rules of oxidation numbers we know that fluorine has an oxidation number of 
. Since there are 
fluorines we multiply 
which equals 
. We can now solve for the oxidation number of Uranium.
, therefore 
has an oxidation number of 
.
Finally, let's look 
. We can once again use the general rule of oxidation numbers that fluorine has a 
oxidation number to simplify this. Therefore 
, therefore Cl has an oxidation number of 
in the product.
Let's now write down the oxidation numbers of all the elements in the products.
Oxidation numbers of elements in products:
Now let's compare the oxidation numbers of the elements in the reactants with those in the products. 
goes from a 
oxidation number to a 
, therefore it is the element that's oxidized (oxidation represents a loss in electrons). 
goes from 
to 
, therefore it is reduced (reduction means gaining electrons). Finally 
's oxidation number is 
in both the products and reactants therefore it is neither oxidized nor reduced. This means that 
is the only element that's oxidized.
Let's start by looking at the equation and assigning oxidation numbers based off of the general rules we are given:
Let's start by looking at the reactants, namely
Next let's look at
so
Now we have the oxidation numbers of all the elements present in this equation that are on the reactant side. I recommend writing down their oxidation numbers next to each other, so the elements that are oxidized and reduced can quickly be determined.
Oxidation numbers of elements in reactants:
Now let's look at the product side of the equation and determine the oxidation numbers of the elements there. First let's look at
Finally, let's look
Let's now write down the oxidation numbers of all the elements in the products.
Oxidation numbers of elements in products:
Now let's compare the oxidation numbers of the elements in the reactants with those in the products.
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In which of the following compounds is the oxidation state of phosphorus the least?
In which of the following compounds is the oxidation state of phosphorus the least?
Let's start by examining each of the compounds we are given case-by-case and applying the general rules of oxidation numbers we know.
Let's start with 
:
We know that based off of the general rules of oxidation states, Oxygen generally has an oxidation number of 
, since we have 
oxygens we must multiply 
to get the total cumulative charge of all the oxygens. This equals 
.
Now we can solve for the oxidation state of phosphorus:
Since we have 
Phosphoruses we multiply our unknown 
by 
.
.
This is equal to 
because the compound has an overall 
net charge.
Solving for 
gives you 
therefore the oxidation state of Phosphorus in this compound is equal to 
.
Next let's looking at 
:
Knowing that an atom in its elemental form has an oxidation state of 
. This compound has an oxidation number of 
.
Now let's look at 
:
Knowing that the oxidation number of hydrogen is 
and that there are 
hydrogens, the total oxidation of all the hydrogens together is 
, therefore our equation is:
so 
, therefore the oxidation state of Phosphorus in this compound is 
.
Now let's consider 
:
Using our elementary rules for the oxidation numbers, we know oxygen has a 
oxidation number. We can also determine the oxidation number of chlorine, using the rule stating that the oxidation number of halogens is usually 
. Since there are three chlorines we must multiply 
, which gives us 
.
Therefore our equation is:
therefore 
, so the oxidation number of phosphorus in this case equals 
.
Finally let's look at 
:
Using the rules stated above for 
, we obtain the equation:
Therefore 
meaning the oxidation number of 
.
This means that the compound with the lowest oxidation state is 
, therefore it is the right answer.
Let's start by examining each of the compounds we are given case-by-case and applying the general rules of oxidation numbers we know.
Let's start with
We know that based off of the general rules of oxidation states, Oxygen generally has an oxidation number of
Now we can solve for the oxidation state of phosphorus:
Since we have
This is equal to
Solving for
Next let's looking at
Knowing that an atom in its elemental form has an oxidation state of
Now let's look at
Knowing that the oxidation number of hydrogen is
so
Now let's consider
Using our elementary rules for the oxidation numbers, we know oxygen has a
Therefore our equation is:
therefore
Finally let's look at
Using the rules stated above for
Therefore
This means that the compound with the lowest oxidation state is
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What species is reduced in the following chemical reaction?
What species is reduced in the following chemical reaction?
Reduction is defined as the gain of electrons so we want to find the element that gains electrons/becomes negatively charged.
Once again our equation is 
Let's start on the reactant side with 
:
Following our general rules of oxidation states, we know that an atom in its elemental form has an oxidation state equal to zero, therefore 
has an oxidation state equal to 
.
Next, consider 
: Based off of our general oxidation state rules we known the oxidation number of oxygen is 
, since we have two oxygens we must multiply 
. This means that the overall combined charge of the oxygens in this compound is equal to 
.
We can now solve for the oxidation state of 
using this equation, which set equal to zero because the net charge of the compound is 
.
, therefore 
Now let's consider 
:
Applying the rule for the oxidation number of oxygen being 
and the rule of hydrogen being equal to +1, we obtain the equation:
where 
is the oxidation number of sulfur.
This simplifies so that 
, therefore the oxidation number of 
.
Moving on to the products, let's consider 
:
We know that the oxidation state of Oxygen is 
based off the rules mentioned above. We can also determine that the oxidation number of 
is equal to 
because lead is most stable when it loses two electrons.
Now we must solve for sulfur's oxidation state:
therefore 
, therefore the oxidation state of sulfur 
.
As for 
, we can just use the general oxidation rules mentioned above to determine 
Looking at the changes of oxidation numbers between elements in the products and reactants we see that there is no change in the oxidation state of sulfur, hydrogen, and oxygen between the products and reactants, but lead goes from 
to 
, so therefore it is reduced.
Reduction is defined as the gain of electrons so we want to find the element that gains electrons/becomes negatively charged.
Once again our equation is
Let's start on the reactant side with
Following our general rules of oxidation states, we know that an atom in its elemental form has an oxidation state equal to zero, therefore
Next, consider
We can now solve for the oxidation state of
Now let's consider
Applying the rule for the oxidation number of oxygen being
where
This simplifies so that
Moving on to the products, let's consider
We know that the oxidation state of Oxygen is
Now we must solve for sulfur's oxidation state:
therefore
As for
Looking at the changes of oxidation numbers between elements in the products and reactants we see that there is no change in the oxidation state of sulfur, hydrogen, and oxygen between the products and reactants, but lead goes from
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What is the oxidation state of copper in 
?
What is the oxidation state of copper in
In order to determine the oxidation state of copper in this compound we first must note the fact that compound has an overall charge of 
.
We can also apply the general rules of oxidation numbers we know to determine the oxidation numbers of hydrogen and oxygen. Since hydrogen is a group 
element, it has an oxidation number of 
. Since there are two hydrogens we must multiple 
which equals 
to get the total charge of both hydrogens.
As for oxygen, based off of our general rules of oxidation numbers, we know that oxygen normally has an oxidation number of 
, so knowing the oxidation numbers of oxygen and hydrogen, and net the charge of the compound, we can setup an equation to solve for the oxidation number of copper as shown below:
therefore 
, therefore copper has an oxidation number equal to 
.
In order to determine the oxidation state of copper in this compound we first must note the fact that compound has an overall charge of
We can also apply the general rules of oxidation numbers we know to determine the oxidation numbers of hydrogen and oxygen. Since hydrogen is a group
As for oxygen, based off of our general rules of oxidation numbers, we know that oxygen normally has an oxidation number of
therefore
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What is the oxidation number of sulfur in 
?
What is the oxidation number of sulfur in
In order to determine the oxidation number we must apply the general rules of oxidation numbers we know. One of these rules states the oxidation number of fluorine is always 
. Since there are 
fluorines that means the fluorines together contribute a 
charge to the compound. Since the overall charge of the compound is 
, we can create an equation with an unknown (charge of sulfur) and solve for it to find the oxidation number of sulfur.
This equation is 
The reason why it is 
instead of just 
is because there are 
sulfurs.
therefore the oxidation number of sulfur is 
.
In order to determine the oxidation number we must apply the general rules of oxidation numbers we know. One of these rules states the oxidation number of fluorine is always
This equation is
therefore the oxidation number of sulfur is
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What is the oxidation state of gold in ?
What is the oxidation state of gold in
Let's first consider the overall compound . Since it is negatively charged overall it is equal to . Applying our elementary rules of oxidation states we know that halogens generally have an oxidation state of , so this means chlorine has an oxidation state of . Since there are chlorines we must multiply to get the charge of all the chlorines together. This is equal to .
We can now solve for oxidation state of gold through creating an equation as shown below:
Simplifying this we get that , therefore the oxidation state of gold is equal to
Let's first consider the overall compound
We can now solve for oxidation state of gold through creating an equation as shown below:
Simplifying this we get that
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Determine the oxidation number of each element in the compound
Determine the oxidation number of each element in the compound
The oxidation number of each element in the compound is:
The oxidation number of chlorine is . There are 3 chlorine atoms present in the compound, so
Because this is a neutral atom, the overall charge is 0. Therefore we can set
Therefore,
We can then solve for the oxidation number for
The oxidation number of each element in the compound
The oxidation number of chlorine is
Because this is a neutral atom, the overall charge is 0. Therefore we can set
Therefore,
We can then solve for the oxidation number for
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Given the following equation, identify the reducing agent.
Given the following equation, identify the reducing agent.
Recall that a reducing agent is being oxidized and is losing electrons. Thus, when looking at the equation, look for which oxidation state is becoming more positive.
Start by assigning oxidation states to the elements.
For the reactants:
has an oxidation state of .
has an oxidation state of .
has an oxidation state of
has an oxidation state of .
Now, assign the oxidation state of the products.
has an oxidation state of .
has an oxidation state of .
has an oxidation state of .
has an oxidation state of .
Only undergoes a loss of electrons. It must be the reducing agent.
Recall that a reducing agent is being oxidized and is losing electrons. Thus, when looking at the equation, look for which oxidation state is becoming more positive.
Start by assigning oxidation states to the elements.
For the reactants:
Now, assign the oxidation state of the products.
Only
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Which compound below has a nitrogen atom at a oxidation state?
Which compound below has a nitrogen atom at a
Let's start by going through each answer case-by-case applying the elementary rules of oxidation states we are given. The most applicable of these rules in this problem are the oxidation state of hydrogen in a compound is generally equal to , and the oxidation state of oxygen in a compound is generally equal to .
Let's start by looking at . We know this compound as a whole is neutrally charged (equal to overall charge), and applying our knowledge of the general oxidation state of hydrogen in a compound and knowing there are hydrogens we know that the hydrogens have a total charge of together, and nitrogen has an unknown oxidation state , so for our equation we have:
solving for gives , giving us the answer we need. Therefore the nitrogen atom in has an oxidation state of , which is the correct answer.
Let's still look at the other cases, however:
As this compound is uncharged we know it's net charge is equal to
Using our knowledge of oxidation states of hydrogen and oxygen and counting the number of hydrogens and oxygen in this compound, we can determine that the total charge of all the hydrogens together is equal to , and that the total charge of the one oxygen is .
Just as we did above we can create an equation and solve for our unknown.
therefore , so the oxidation state of nitrogen in is . This means this isn't the right compound.
Next let's look at :
We can quickly apply our knowledge of oxidation states of oxygen knowing that the oxidation state of oxygen in compounds is generally . Since there are oxygens we must multiply to find out the total charge contributed by all the oxygens, which is .
So our equation for this becomes:
therefore and the oxidation state of nitrogen is , which isn't the answer we are looking for.
Finally the last compound is :
With this compound we can apply the rule we know about hydrogen's oxidation state in a compound being equal to . Since there are hydrogens, we must multiply which equals .
Since there are two nitrogens, our equation is:
and . Therefore the oxidation state of nitrogen is equal to .
Let's start by going through each answer case-by-case applying the elementary rules of oxidation states we are given. The most applicable of these rules in this problem are the oxidation state of hydrogen in a compound is generally equal to
Let's start by looking at
solving for
Let's still look at the other cases, however:
As this compound is uncharged we know it's net charge is equal to
Using our knowledge of oxidation states of hydrogen and oxygen and counting the number of hydrogens and oxygen in this compound, we can determine that the total charge of all the hydrogens together is equal to
Just as we did above we can create an equation and solve for our unknown.
therefore
Next let's look at
We can quickly apply our knowledge of oxidation states of oxygen knowing that the oxidation state of oxygen in compounds is generally
So our equation for this becomes:
therefore
Finally the last compound is
With this compound we can apply the rule we know about hydrogen's oxidation state in a compound being equal to
Since there are two nitrogens, our equation is:
and
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In the given reaction, which element(s) is/are oxidized?
In the given reaction, which element(s) is/are oxidized?
Let's start by looking at the equation and assigning oxidation numbers based off of the general rules we are given:
Let's start by looking at the reactants, namely . Knowing that an atom in its elemental form has an oxidation number of , has an oxidation number equal to .
Next let's look at . This is a neutral compound so know the oxidation number of the whole compound must equal . Going off of our general rules, the oxidation number of fluorine is equal . Since we have 3 fluorines we multiply the oxidation number by to get the cumulative oxidation number of all the fluorines together to get . So now we must solve for the oxidation of Cl, which is unknown. This is done through a simple algebraic equation:
so . Therefore the oxidation number of Chlorine is .
Now we have the oxidation numbers of all the elements present in this equation that are on the reactant side. I recommend writing down their oxidation numbers next to each other, so the elements that are oxidized and reduced can quickly be determined.
Oxidation numbers of elements in reactants:
Now let's look at the product side of the equation and determine the oxidation numbers of the elements there. First let's look at . Knowing the general rules of oxidation numbers we know that fluorine has an oxidation number of . Since there are fluorines we multiply which equals . We can now solve for the oxidation number of Uranium.
, therefore has an oxidation number of .
Finally, let's look . We can once again use the general rule of oxidation numbers that fluorine has a oxidation number to simplify this. Therefore
, therefore Cl has an oxidation number of in the product.
Let's now write down the oxidation numbers of all the elements in the products.
Oxidation numbers of elements in products:
Now let's compare the oxidation numbers of the elements in the reactants with those in the products. goes from a oxidation number to a , therefore it is the element that's oxidized (oxidation represents a loss in electrons). goes from to , therefore it is reduced (reduction means gaining electrons). Finally 's oxidation number is in both the products and reactants therefore it is neither oxidized nor reduced. This means that is the only element that's oxidized.
Let's start by looking at the equation and assigning oxidation numbers based off of the general rules we are given:
Let's start by looking at the reactants, namely
Next let's look at
so
Now we have the oxidation numbers of all the elements present in this equation that are on the reactant side. I recommend writing down their oxidation numbers next to each other, so the elements that are oxidized and reduced can quickly be determined.
Oxidation numbers of elements in reactants:
Now let's look at the product side of the equation and determine the oxidation numbers of the elements there. First let's look at
Finally, let's look
Let's now write down the oxidation numbers of all the elements in the products.
Oxidation numbers of elements in products:
Now let's compare the oxidation numbers of the elements in the reactants with those in the products.
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In which of the following compounds is the oxidation state of phosphorus the least?
In which of the following compounds is the oxidation state of phosphorus the least?
Let's start by examining each of the compounds we are given case-by-case and applying the general rules of oxidation numbers we know.
Let's start with :
We know that based off of the general rules of oxidation states, Oxygen generally has an oxidation number of , since we have oxygens we must multiply to get the total cumulative charge of all the oxygens. This equals .
Now we can solve for the oxidation state of phosphorus:
Since we have Phosphoruses we multiply our unknown by .
.
This is equal to because the compound has an overall net charge.
Solving for gives you therefore the oxidation state of Phosphorus in this compound is equal to .
Next let's looking at :
Knowing that an atom in its elemental form has an oxidation state of . This compound has an oxidation number of .
Now let's look at :
Knowing that the oxidation number of hydrogen is and that there are hydrogens, the total oxidation of all the hydrogens together is , therefore our equation is:
so , therefore the oxidation state of Phosphorus in this compound is .
Now let's consider :
Using our elementary rules for the oxidation numbers, we know oxygen has a oxidation number. We can also determine the oxidation number of chlorine, using the rule stating that the oxidation number of halogens is usually . Since there are three chlorines we must multiply , which gives us .
Therefore our equation is:
therefore , so the oxidation number of phosphorus in this case equals .
Finally let's look at :
Using the rules stated above for , we obtain the equation:
Therefore meaning the oxidation number of .
This means that the compound with the lowest oxidation state is , therefore it is the right answer.
Let's start by examining each of the compounds we are given case-by-case and applying the general rules of oxidation numbers we know.
Let's start with
We know that based off of the general rules of oxidation states, Oxygen generally has an oxidation number of
Now we can solve for the oxidation state of phosphorus:
Since we have
This is equal to
Solving for
Next let's looking at
Knowing that an atom in its elemental form has an oxidation state of
Now let's look at
Knowing that the oxidation number of hydrogen is
so
Now let's consider
Using our elementary rules for the oxidation numbers, we know oxygen has a
Therefore our equation is:
therefore
Finally let's look at
Using the rules stated above for
Therefore
This means that the compound with the lowest oxidation state is
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What species is reduced in the following chemical reaction?
What species is reduced in the following chemical reaction?
Reduction is defined as the gain of electrons so we want to find the element that gains electrons/becomes negatively charged.
Once again our equation is
Let's start on the reactant side with :
Following our general rules of oxidation states, we know that an atom in its elemental form has an oxidation state equal to zero, therefore has an oxidation state equal to .
Next, consider : Based off of our general oxidation state rules we known the oxidation number of oxygen is , since we have two oxygens we must multiply . This means that the overall combined charge of the oxygens in this compound is equal to .
We can now solve for the oxidation state of using this equation, which set equal to zero because the net charge of the compound is .
, therefore
Now let's consider :
Applying the rule for the oxidation number of oxygen being and the rule of hydrogen being equal to +1, we obtain the equation:
where is the oxidation number of sulfur.
This simplifies so that , therefore the oxidation number of .
Moving on to the products, let's consider :
We know that the oxidation state of Oxygen is based off the rules mentioned above. We can also determine that the oxidation number of is equal to because lead is most stable when it loses two electrons.
Now we must solve for sulfur's oxidation state:
therefore , therefore the oxidation state of sulfur .
As for , we can just use the general oxidation rules mentioned above to determine
Looking at the changes of oxidation numbers between elements in the products and reactants we see that there is no change in the oxidation state of sulfur, hydrogen, and oxygen between the products and reactants, but lead goes from to , so therefore it is reduced.
Reduction is defined as the gain of electrons so we want to find the element that gains electrons/becomes negatively charged.
Once again our equation is
Let's start on the reactant side with
Following our general rules of oxidation states, we know that an atom in its elemental form has an oxidation state equal to zero, therefore
Next, consider
We can now solve for the oxidation state of
Now let's consider
Applying the rule for the oxidation number of oxygen being
where
This simplifies so that
Moving on to the products, let's consider
We know that the oxidation state of Oxygen is
Now we must solve for sulfur's oxidation state:
therefore
As for
Looking at the changes of oxidation numbers between elements in the products and reactants we see that there is no change in the oxidation state of sulfur, hydrogen, and oxygen between the products and reactants, but lead goes from
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What is the oxidation state of copper in ?
What is the oxidation state of copper in
In order to determine the oxidation state of copper in this compound we first must note the fact that compound has an overall charge of .
We can also apply the general rules of oxidation numbers we know to determine the oxidation numbers of hydrogen and oxygen. Since hydrogen is a group element, it has an oxidation number of . Since there are two hydrogens we must multiple which equals to get the total charge of both hydrogens.
As for oxygen, based off of our general rules of oxidation numbers, we know that oxygen normally has an oxidation number of , so knowing the oxidation numbers of oxygen and hydrogen, and net the charge of the compound, we can setup an equation to solve for the oxidation number of copper as shown below:
therefore , therefore copper has an oxidation number equal to .
In order to determine the oxidation state of copper in this compound we first must note the fact that compound has an overall charge of
We can also apply the general rules of oxidation numbers we know to determine the oxidation numbers of hydrogen and oxygen. Since hydrogen is a group
As for oxygen, based off of our general rules of oxidation numbers, we know that oxygen normally has an oxidation number of
therefore
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What is the oxidation number of sulfur in ?
What is the oxidation number of sulfur in
In order to determine the oxidation number we must apply the general rules of oxidation numbers we know. One of these rules states the oxidation number of fluorine is always . Since there are fluorines that means the fluorines together contribute a charge to the compound. Since the overall charge of the compound is , we can create an equation with an unknown (charge of sulfur) and solve for it to find the oxidation number of sulfur.
This equation is The reason why it is instead of just is because there are sulfurs.
therefore the oxidation number of sulfur is .
In order to determine the oxidation number we must apply the general rules of oxidation numbers we know. One of these rules states the oxidation number of fluorine is always
This equation is
therefore the oxidation number of sulfur is
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Determine the oxidation number of each element in the compound
Determine the oxidation number of each element in the compound
The oxidation number of each element in the compound is:
The oxidation number of chlorine is . There are 3 chlorine atoms present in the compound, so
Because this is a neutral atom, the overall charge is 0. Therefore we can set
Therefore,
We can then solve for the oxidation number for
The oxidation number of each element in the compound
The oxidation number of chlorine is
Because this is a neutral atom, the overall charge is 0. Therefore we can set
Therefore,
We can then solve for the oxidation number for
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Given the following equation, identify the reducing agent.
Given the following equation, identify the reducing agent.
Recall that a reducing agent is being oxidized and is losing electrons. Thus, when looking at the equation, look for which oxidation state is becoming more positive.
Start by assigning oxidation states to the elements.
For the reactants:
has an oxidation state of .
has an oxidation state of .
has an oxidation state of
has an oxidation state of .
Now, assign the oxidation state of the products.
has an oxidation state of .
has an oxidation state of .
has an oxidation state of .
has an oxidation state of .
Only undergoes a loss of electrons. It must be the reducing agent.
Recall that a reducing agent is being oxidized and is losing electrons. Thus, when looking at the equation, look for which oxidation state is becoming more positive.
Start by assigning oxidation states to the elements.
For the reactants:
Now, assign the oxidation state of the products.
Only
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