Redox Reactions - College Chemistry

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.

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 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 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 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

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.

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 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 .

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

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.

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.

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.

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 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 .

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.

The reducing agent is the compound which contains the element being oxidized. Oxidation is defined as the loss of electrons, so the element which is oxidized becomes more positively charged.

In order to determine which element is oxidized we need to calculate the oxidation numbers of all the elements in the reactants and products and compare them. The element which has an oxidation number that increases overall is the one that is oxidized. We determine the oxidation numbers by applying the elementary rules of oxidation numbers to each compound.

For reference the chemical equation is:

Let's start with the reactants:

because a rule of oxidation states is that oxidation number of all halogens is generally .

for the same reason as bromine.

because is neutrally charged and bromine has a charge of and there are two bromines, so in order for the compound to maintain a neutral charge iron must have a charge of .

for the same why iron is, except in this case it is due to chlorine's charge instead of bromine.

Products:

because the oxidation number of all halogens is generally .

for the same reason as bromine.

and because they both must be numbers that allow each of their respective compounds to be neutrally charged. For the same reasoning as in the reactants they both must be .

Since there are no changes in the oxidation numbers of any of these elements, this isn't a redox reaction therefore there is no reducing agent.

The oxidizing agent is the compound which contains the element being reduced. Reduction is defined as gaining of electrons, so the element which is reduced becomes more negatively charged.

In order to determine which element is reduced we need to calculate the oxidation numbers of all the elements in the reactants and products and compare them. The element which has an oxidation number that decreases overall is the one that is reduced and therefore the oxidizing agent. We determine the oxidation numbers by applying the elementary rules of oxidation numbers to each compound.

For reference the chemical equation is:

Let's start with the reactants:

because it is in the same group as oxygen which by the general rules of oxidation numbers has a charge of . Although sulfur can assume other oxidation numbers, it is also in this particular case because chromium is a transition metal that prefers to lose electrons to become more stable (become more positively charged) and since sulfur has a greater electronegativity it prefers a negatively charged state in this case.

because sulfur is and since is a neutral compound chromium must be

because a general rule of oxidation numbers is that oxygen usually has an oxidation number of .

In order to determine what phosphorus is we must look at the phosphate ion . Phosphate is an anion which has a charge of and in order for that to be possible with four oxygens () P must have a charge of .

= because the phosphate anion has a charge of , so must offset that charge for to be neutrally charged.

Products:

for the same reason it is in the reactants, except because is the transition metal in this case which prefers to lose electrons.

because is neutrally charged, so it must offset the charge of .

for the same reason as in the reactants

for the same reason in the reactants

to offset the charge of the phosphate compound it is attached to.

Now let's compare the oxidation numbers of each element:

Oxygen, phosphate, and sulfur have the same oxidation states in both products and reactants.

Chromium increases from an oxidation state of to .

Finally, manganese decreases from an oxidation state of to , so it is the compound which is reduced because it gains electrons. Therefore the compound containing manganese is the oxidizing agent. This compound is , which is the right answer.

For this question, we're given a reduction-oxidation (redox) reaction. We're asked to identify an answer choice that correctly states what the reducing agent is and what the oxidizing agent is.

Notice that another way we can write this reaction is by breaking up the ions as if they were dissociated in solution.

Notice in the above example that the potassium ion remains in the same oxidation state. This is what is known as a spectator ion, meaning that it doesn't really participate in the reaction and remains unchanged on reactant and product sides of the equation. Hence, we can remove it from both sides to simplify things.

In the simplified reaction shown above, we can see that the chlorine element is being reduced, while the iodine is being oxidized. Since the chlorine is being reduced, the iodine must be acting as the agent responsible for this. Hence, the is the reducing agent. Moreover, since the iodine is being oxidized, the chlorine must be the agent responsible for this. Thus, is the oxidizing agent.

Remember that the we're referring to as the reducing agent is originally in the form of , which we can thus state as the reducing agent in the overall, unsimplified reaction.

For this question, we're given a redox reaction occurring within a voltaic cell. We're also supplied with the reduction potentials for the elements in the reaction, and are asked to find the standard free energy change as well as the equilibrium constant of the reaction at a given temperature.

Recall that in a voltaic cell, the reaction does not consume energy but rather produces it. The anode is where oxidation occurs, while the cathode is where reduction takes place. Based on the reaction given to us in the question stem, the anode will consist of solid , while the cathode will consist of solid . As oxidation occurs in the anode, the electrons liberated from the oxidation of solid will result in the production of , which will exist as a cation in the aqueous solution. Moreover, these electrons will flow spontaneously from the anode to the cathode, releasing energy in the process. When the cathode receives these electrons, the in the aqueous solution will be reduced and thus deposited onto the cathode terminal as solid .

Knowing this, we can calculate for the reaction.

Now that we have for the reaction, we can plug this into the Nernst equation to solve for the equilibrium constant.

A couple things to note. The value for used in the equation above is because this is the number of electrons transferred in the balanced reaction. Also, note that the value for this equilibrium expression is huge, meaning that the reaction is driven far to the right, making it spontaneous and capable of releasing a great deal of energy.

To find the value of , there are a few ways we can go about it. We can use the equilibrium value we just obtained, or we can use the value.

Using :

Using :

As shown above, either method leads us to the same value for . And once again, note how large of a negative value this term is. This is consistent with the large value for the equilibrium constant. The reaction equilibrium lies far to the right and produces an enormous amount of energy.