Reduction always takes place at the cathode. This means that a substance is receiving electrons.

In the reaction, the gold ions are receiving electrons in order to create gold atoms. This takes place at the cathode.

Copper ions and gold are products, and will not be oxidized or reduced. Copper is reduced at the anode during this reaction.

There are two concepts to consider in this problem.  First, the question asks for the .  Reaction 3 has  being reduced so the potential for the half reaction becomes negative.  half reaction appears the same in reaction 3 so the potential is the same. Second, negative voltages indicate non-spontaneity and positive voltages are spontaneous.  

When one molecule is converted into two or more smaller molecules, the reaction is considered a decomposition reaction. In this example, none of the elements under go a change in oxidation number, so this is not an oxidation-reduction reaction. (Calcium is always , oxygen is always , and carbon remains  throughout the reaction.)  

Combustion reactions always produce  and excess heat. Depending on the molecule that is oxidized by oxygen, which in this case was methane , the ratios between  and  can vary.

Potassium is a Group I element, so to get to a filled valence shell, it will lost one electron, yielding .

Arsenic is a Group 5 element, so it needs to gain three electrons to obtain a filled valence shell, yielding .

In order to balance out the charges, the resultant salt will be .

First, we must know what ferrous sulfate is. Ferrous refers to , and sulfate has the formula . When we combine the two together we get .

Calcium is a divatent cation and iodide is a monovalent anion, so their salt is . The ion exchange reaction is then:

The net ionic equation is derived by removing all spectator ions from the total ionic equation (in which all ions are listed). To put it another way, the net ionic equation involves only the ions that participate in a reaction which, in this case, is the precipitation of barium sulfate.

Begin by writing all aqueous compounds in their dissociated (ionic) forms.

Cancel out any ions that appear in equal quantities on both sides of the equation. In this case, we can cancel the nitrate and potassium ions.

This is our net ionic equation.

Combustion is the chemical reaction of a hydrocarbon with molecular oxygen, and it always produces carbon dioxide and water. Knowing the reactants and products, the unbalanced equation must be: 

We start by balancing the hydrogens. Since there are 10 on the left and only 2 on the right, we put a coefficient of 5 on water.

Similarly, we balance carbons by putting a 4 on the carbon dioxide.

To find the number of oxygens on the right, we multiply the 4 coefficient by the 2 subscript on O (which gets us 8 oxygens) and then add the 5 oxygens from the 5 water molecules to get a total of 13. The needed coefficient for  on the left would then have to be 13/2.

Because fractional coefficients are not allowed, we mutiply every coefficient by 2 to find our final reaction:

When we check the activity series, it is fairly easy to see that aluminum metal is more reactive than zinc metal. So, in this case, the two metals undergo a redox reaction, where the aqueous  is reduced to solid , and the solid  is oxidized to aqueous . These charges are the common oxidation states for zinc and aluminum and should be memorized.

Because  is the new species, it bonds with 3  ions. The unbalanced equation is:

We note that there are 2 chlorine atoms on the left and 3 chlorine atoms on the right. To balance, we use a 3 coefficient on the left and a 2 coefficient on the right. This gives a total of 6 chlorine atoms on eahc side.

However, now we have also increased the amounts of zinc and aluminum. We copy the necessary coefficients to balance those—2 for aluminum on the left, 3 for zinc on the right—and we are done:

To find the net ionic form of this chemical equation, first balance the equation, then dissociate all soluble components. In other words, separate the aqueous terms into two separate molecules/atoms. For example,  dissociates into  and . After separating all soluble terms, balance the equation and cancel out any terms that repeat on both sides of the equation. For example,  will be on both sides of the equation after separating all terms so you can cross them out. Below are the steps taken to find the net ionic form of this chemical equation.

Write out the dissociated forms of each species, if applicable.

Cross out species that appear on both sides of the equation. The net ionic equation is: