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 .

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.

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.

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