Reactions - AP Chemistry
Card 0 of 1354
You are in chemistry lab performing a titration. You were given 15 mL of an aqueous solution with an unknown concentration of acetic acid, 
to solve through titration with concentrated sodium hydroxide, 
. You know that the pKa of acetic acid is 4.75 and that your titrant is 0.1 M sodium hydroxide, 
.
The endpoint was determined at 10 mL of sodium hydroxide, 
. What is the pH after 5 mL of 
was added?
You are in chemistry lab performing a titration. You were given 15 mL of an aqueous solution with an unknown concentration of acetic acid,
The endpoint was determined at 10 mL of sodium hydroxide,
At the half end point, the 
. This can be determined by the Henderson-Hasselbalch equation if it is not clear.
Since the endpoint of the titration is that there are 10 mL of 0.1 M NaOH added, that means that there are 0.001 moles of acetic acid.
When 5 mL of NaOH is added, there are 0.0005 moles of acetic acid and 0.0005 moles of acetate formed.
Therefore pH= pKa
At the half end point, the
Since the endpoint of the titration is that there are 10 mL of 0.1 M NaOH added, that means that there are 0.001 moles of acetic acid.
When 5 mL of NaOH is added, there are 0.0005 moles of acetic acid and 0.0005 moles of acetate formed.
Therefore pH= pKa
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Determine which of these solution combinations form a buffer.
Determine which of these solution combinations form a buffer.
First to go through why the other ones are wrong:
Strong base + strong acid neutralizes and does not form a buffer solution
Strong base + weak base does not form a buffer - would need an acid
Strong base + weak acid = all weak acid converted to conjugate base
The correct answer is:
Strong base + weak acid = half converted to conjugate base with half leftover as weak acid, with all the components for a buffer
First to go through why the other ones are wrong:
Strong base + strong acid neutralizes and does not form a buffer solution
Strong base + weak base does not form a buffer - would need an acid
Strong base + weak acid = all weak acid converted to conjugate base
The correct answer is:
Strong base + weak acid = half converted to conjugate base with half leftover as weak acid, with all the components for a buffer
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Determine which combination of solutions would create a buffer solution.
Determine which combination of solutions would create a buffer solution.
For all the other options there is no ammonium leftover with which to serve as the weak acid in the buffer system, the ammonium is all used up and converted to ammonia. However in the correct answer choice, there is enough ammonium leftover after the reaction with the sodium hydroxide.
For all the other options there is no ammonium leftover with which to serve as the weak acid in the buffer system, the ammonium is all used up and converted to ammonia. However in the correct answer choice, there is enough ammonium leftover after the reaction with the sodium hydroxide.
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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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Balance the following redox reaction in acidic conditions:
Balance the following redox reaction in acidic conditions:
Start by writing the half reactions for the equation
All atoms except for O and H are already balanced, so we can go straight to balancing O and H atoms. First balance O atoms by adding H2O
Now balance H atoms with H+
Balance each half reaction's electrical charge with electrons
Now combine both half reactions to get the overall reaction. Be sure to cancel out electrons by multiplying the oxidation reaction by 5 and the reduction reaction by 2. This will give 10 e- on each side of the overall reaction so that they cancel out.
Overall Reaction:
Notice that we have H2O and H+ on both sides of the equation, so we need to simplify. Subtract the 5 H2O on the reactant side from the 8 H2O on the product side which leaves 3 H2O on the reactant side. Then subtract the 10 H+ on the product side from the 16 H+ on the reactant side which leave 6 H+ on the reactant side. Now we have the simplified overall reaction
Start by writing the half reactions for the equation
All atoms except for O and H are already balanced, so we can go straight to balancing O and H atoms. First balance O atoms by adding H2O
Now balance H atoms with H+
Balance each half reaction's electrical charge with electrons
Now combine both half reactions to get the overall reaction. Be sure to cancel out electrons by multiplying the oxidation reaction by 5 and the reduction reaction by 2. This will give 10 e- on each side of the overall reaction so that they cancel out.
Overall Reaction:
Notice that we have H2O and H+ on both sides of the equation, so we need to simplify. Subtract the 5 H2O on the reactant side from the 8 H2O on the product side which leaves 3 H2O on the reactant side. Then subtract the 10 H+ on the product side from the 16 H+ on the reactant side which leave 6 H+ on the reactant side. Now we have the simplified overall reaction
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Which combination(s) would create a buffer solution?
I. Weak acid
II. Weak acid's conjugate base
III. Strong acid
IV. Strong base
V. Weak base
VI. Weak base's conjugate acid
Which combination(s) would create a buffer solution?
I. Weak acid
II. Weak acid's conjugate base
III. Strong acid
IV. Strong base
V. Weak base
VI. Weak base's conjugate acid
A buffer solution is formed from the equilibrium of a weak acid and its conjugate base, or from a weak base and its conjugate acid. It's ability to "buffer" the pH or keep it from changing in large amounts in from the switching between these two forms weak and its conjugate.
A buffer solution is formed from the equilibrium of a weak acid and its conjugate base, or from a weak base and its conjugate acid. It's ability to "buffer" the pH or keep it from changing in large amounts in from the switching between these two forms weak and its conjugate.
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Suppose that the only elements a compound contains are carbon, hydrogen, and oxygen. If complete combustion of 
of this compound produces 
of 
and 
of 
, what is this compound's molecular formula?
Note: The molecular weight of this compound is 
.
Suppose that the only elements a compound contains are carbon, hydrogen, and oxygen. If complete combustion of
Note: The molecular weight of this compound is
To answer this question, it's important to remember the details of combustion reactions. In the presence of oxygen and a sufficient amount of energy to initiate the process, combustion reactions that go to completion will result in the complete oxidation and breakdown of the initial reactant.
The 
produced from the reaction contains all of the carbon in the original reactant. Likewise, the 
produced contains all the hydrogen in the original compound. The oxygen content in the original compound is everything left over.
The first step then is to use the mass of each product to find the mass of each element in the original compound. To do this, we'll need to use the molar mass for each of the products as well as the individual elements.
The combined mass of carbon and hydrogen in the original compound is 
.
The remaining mass will come from oxygen. Thus, the original compound will contain 
of oxygen.
Once we have the mass of each element, we can use that information to calculate the number of moles of each element in the original compound.
Now that we have the relative molar amounts of each element in the original compound, we can find the empirical formula by finding the smallest whole number ratios.
From this, we can conclude that the empirical mass of the compound is 
.
Finally, since we are given the compound's molecular weight, we can use that information to find the molecular formula for the compound.
To answer this question, it's important to remember the details of combustion reactions. In the presence of oxygen and a sufficient amount of energy to initiate the process, combustion reactions that go to completion will result in the complete oxidation and breakdown of the initial reactant.
The
The first step then is to use the mass of each product to find the mass of each element in the original compound. To do this, we'll need to use the molar mass for each of the products as well as the individual elements.
The combined mass of carbon and hydrogen in the original compound is
The remaining mass will come from oxygen. Thus, the original compound will contain
Once we have the mass of each element, we can use that information to calculate the number of moles of each element in the original compound.
Now that we have the relative molar amounts of each element in the original compound, we can find the empirical formula by finding the smallest whole number ratios.
From this, we can conclude that the empirical mass of the compound is
Finally, since we are given the compound's molecular weight, we can use that information to find the molecular formula for the compound.
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Electrolysis of an unknown metal chloride, 
, with a current of 
for 
seconds deposits 
of the metal at the cathode. What is the metal?
Electrolysis of an unknown metal chloride,
Start by writing out the equation that illustrates the plating out of the metal:
Note the stoichiometric ratio for the moles of electrons to the moles of the metal.
Next, recall the following equation:
, where 
is the charge, 
is the current, and 
is the time.
Plug in the given information to find 
.
Next, recall Faraday's law of electrolysis that equates Faraday's constant and the amount of electrons together to find the charge needed to deposit one mole of a particular substance.
Thus, we can set up the following to take the charge of the electrolysis through to figure out the number of moles unknown metal that was plated out.
Now, since we have the number of grams of the metal deposited, we can find the molar mass of the unknown metal.
The molar mass indicates that the metal must be titanium.
Start by writing out the equation that illustrates the plating out of the metal:
Note the stoichiometric ratio for the moles of electrons to the moles of the metal.
Next, recall the following equation:
Plug in the given information to find
Next, recall Faraday's law of electrolysis that equates Faraday's constant and the amount of electrons together to find the charge needed to deposit one mole of a particular substance.
Thus, we can set up the following to take the charge of the electrolysis through to figure out the number of moles unknown metal that was plated out.
Now, since we have the number of grams of the metal deposited, we can find the molar mass of the unknown metal.
The molar mass indicates that the metal must be titanium.
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Scandium can be plated out of a solution containing 
according to the following half reaction:
How many minutes would it take to plate out 
of scandium using a current of 
?
Scandium can be plated out of a solution containing
How many minutes would it take to plate out
Recall the following formula:
, where 
is the charge in Coulombs, 
is the current in Amps, 
is the time in seconds, 
is the number of electrons, and 
is Faraday's constant.
Start by finding how many moles of scandium was plated out.
Now, use the stoichiometric ratio present in the half reaction to find the number of moles of electrons.
Now, use the above formula to solve for the time.
Now, convert the seconds to minutes. Remember that your answer should only have 
significant figures.
Recall the following formula:
Start by finding how many moles of scandium was plated out.
Now, use the stoichiometric ratio present in the half reaction to find the number of moles of electrons.
Now, use the above formula to solve for the time.
Now, convert the seconds to minutes. Remember that your answer should only have
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Consider the following redox reaction.
If 
has a reduction potential of 
and 
has a reduction potential of 
, what is the 
for this redox reaction?
Consider the following redox reaction.
If
In this question, we're given an overall redox reaction as well as the relevant reduction potentials, and we're asked to solve for 
.
To begin with, we can show each of the individual half reactions to make it more clear.
For gold:
For bromine:
Note that the reduction potential for each of these half reactions is positive. This means that for both elements, their reduction is spontaneous. The more positive the voltage (or the less negative), the more spontaneous it is.
From the overall question given to us in the question stem, we see that bromine is not being reduced but, rather, it is being oxidized. Since oxidation is the reverse of reduction, the reduction potential maintains the same magnitude but the sign in front of it changes. Thus, the oxidation of bromine has a value of 
. Furthermore, because the overall reaction shows gold being reduced, we don't need to change the sign of gold's reduction potential.
To find the overall 
for the reaction, we simply just add these two values together.
One very important thing to note is that we did not need to multiply the reduction potential for either of the half reactions. Even though the reduction reaction for gold needs two stoichiometric equivalents, and bromine's oxidation needs three stoichiometric equivalents, the values of 
and 
do not change. This is because the any 
value represents an intrinsic property. In other words, the value is not dependent on the amount of material present. As you add more material, it is true that there will be greater electron flow. But at the same time, there will also be more energy change as these electrons flow. The consequence is that both of these values (electron flow and change in energy) change proportionately, such that their ratio will always equal the value of 
that is characteristic of that redox reaction.
In this question, we're given an overall redox reaction as well as the relevant reduction potentials, and we're asked to solve for
To begin with, we can show each of the individual half reactions to make it more clear.
For gold:
For bromine:
Note that the reduction potential for each of these half reactions is positive. This means that for both elements, their reduction is spontaneous. The more positive the voltage (or the less negative), the more spontaneous it is.
From the overall question given to us in the question stem, we see that bromine is not being reduced but, rather, it is being oxidized. Since oxidation is the reverse of reduction, the reduction potential maintains the same magnitude but the sign in front of it changes. Thus, the oxidation of bromine has a value of
To find the overall
One very important thing to note is that we did not need to multiply the reduction potential for either of the half reactions. Even though the reduction reaction for gold needs two stoichiometric equivalents, and bromine's oxidation needs three stoichiometric equivalents, the values of
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The standard reduction potential of silver is equal to 
. If silver acts as the anode in a voltaic cell, which of the following would be able to act as the cathode?
The standard reduction potential of silver is equal to
In this question, we're given the standard reduction potential of silver and told that it acts as the anode in a voltaic cell. We're then asked to identify a compound that could act as the cathode.
First, there are a few things we need to recognize. For one thing, we're told this is a voltaic cell, which hosts spontaneous redox reactions. This means that energy will be liberated from the reaction and, thus, the overall potential of the cell needs to be positive.
Moreover, the anode is the half-cell where oxidation occurs. In this case, since silver is serving as the anode, it is silver that will be oxidized. The electrons released from this oxidation will travel through the wire to the cathode, liberating energy available to do work in the process. Once at the cathode, the electron will be used to reduce whichever compound is at the cathode.
Since we're given the standard reduction potential of silver and we know the overall reaction must be spontaneous, we can determine what kind of reduction potential the cathode would need to have. Remember that in an electrochemical cell, the cell potential can be expressed as follows.
Moreover, as was mentioned previously, since this is a voltaic cell the reaction will be spontaneous and thus the 
value must be positive.
Plugging in the value for 
gives us the following.
This shows us that the cathode half-cell needs to have a compound with a standard reduction potential that is greater than silver's. The only answer choice that fits this criteria is cobalt, which has a standard reduction potential of 
.
In this question, we're given the standard reduction potential of silver and told that it acts as the anode in a voltaic cell. We're then asked to identify a compound that could act as the cathode.
First, there are a few things we need to recognize. For one thing, we're told this is a voltaic cell, which hosts spontaneous redox reactions. This means that energy will be liberated from the reaction and, thus, the overall potential of the cell needs to be positive.
Moreover, the anode is the half-cell where oxidation occurs. In this case, since silver is serving as the anode, it is silver that will be oxidized. The electrons released from this oxidation will travel through the wire to the cathode, liberating energy available to do work in the process. Once at the cathode, the electron will be used to reduce whichever compound is at the cathode.
Since we're given the standard reduction potential of silver and we know the overall reaction must be spontaneous, we can determine what kind of reduction potential the cathode would need to have. Remember that in an electrochemical cell, the cell potential can be expressed as follows.
Moreover, as was mentioned previously, since this is a voltaic cell the reaction will be spontaneous and thus the
Plugging in the value for
This shows us that the cathode half-cell needs to have a compound with a standard reduction potential that is greater than silver's. The only answer choice that fits this criteria is cobalt, which has a standard reduction potential of
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Which compound is the reducing agent in this reaction?:
Which compound is the reducing agent in this reaction?:
A reducing agent is the compound in the reactants which becomes oxidized.
Oxidation is defined as the loss of electrons and is shown by the element changing to have a more positive oxidation number in the reaction.
To determine the reducing agent we simply see which compound is oxidized by determining the oxidation numbers of the elements in both the reactants and products, and comparing them.
Reactants:
For 
its oxidation state is simply zero because the oxidation number of an atom in its elemental form is always zero.
Same goes 
, which is why its oxidation number is also zero.
Products:
has a charge of 
, so its oxidation number is 
.
has a charge of 
, so its oxidation number is 
Results:
Since 
loses electrons it is oxidized and therefore its the reducing agent.
None of the products can be the reducing agent because this reaction isn't reversible.
A reducing agent is the compound in the reactants which becomes oxidized.
Oxidation is defined as the loss of electrons and is shown by the element changing to have a more positive oxidation number in the reaction.
To determine the reducing agent we simply see which compound is oxidized by determining the oxidation numbers of the elements in both the reactants and products, and comparing them.
Reactants:
For
Same goes
Products:
Results:
Since
None of the products can be the reducing agent because this reaction isn't reversible.
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