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

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 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.

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.

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.

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.

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.

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.

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

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