Balancing redox reactions involves the following steps:

1. Divide the reaction into oxidation and reduction half reactions and balance all atoms that are not oxygen and hydrogen:

$\displaystyle SO_{2} \rightarrow SO_{4}^{2-}$

$\displaystyle Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+}$

2. Balance the oxygens by adding water molecules on the opposite side of the reactions:

$\displaystyle 2H_{2}O + SO_{2} \rightarrow SO_{4}^{2-}$

$\displaystyle Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$

3. Balance the hydrogens by adding protons to the opposite side of the equation:

$\displaystyle 2H_{2}O + SO_{2} \rightarrow SO_{4}^{2-} + 4H^{+}$

$\displaystyle 14H^{+} + Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$

4. Add electrons in order to equal the charges on both sides of the equation:

$\displaystyle 2H_{2}O + SO_{2} \rightarrow SO_{4}^{2-} + 4H^{+} + 2e^{-}$

$\displaystyle 6e^{-} + 14H^{+} + Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$

5. In order to make the electron exchange equal in each half step, we must multiply the top half reaction by 3:

 $\displaystyle 6H_{2}O + 3SO_{2} \rightarrow 3SO_{4}^{2-} + 12H^{+} + 6e^{-}$

$\displaystyle 6e^{-} + 14H^{+} + Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$

6. Add up the reactants and products while cancelling out substances on opposite sides of the reactions. For example: we will cancel the 6 water molecules as reactants and be left with only one water molecule as a product. In addition, only 2 protons will be left on the reactant's side after canceling the 12 from the product's side.

$\displaystyle Cr_{2}O_{7}^{2-} + 3SO_{2} + 2H^{+} \rightarrow 2Cr^{3+} + 3SO_{4}^{2-} + H_{2}O$

In the balanced redox reaction, the coefficient on sulfur dioxide is 3.

To begin, we will need to separate the given reaction into the two half-reactions by identifying changes in oxidation number. In this case, mercury (Hg) and phosphorus (P) show a change in oxidation number. Mercury begins with an oxidation number of zero, and ends with an oxidation number of $\displaystyle +1$. Phosphorus begins with an oxidation number of $\displaystyle +3$ and ends with an oxidation number of $\displaystyle +2$. Note that the oxidation numbers for fluorine and iodine reamain constant at $\displaystyle -1$ for each.

Now we can begin to look at the half-reactions.

$\displaystyle PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$\displaystyle Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Balance the atoms.

$\displaystyle 2PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$\displaystyle 2Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Now balance the electrons. We know that each mercury atom loses one electron and each fluorine atom gains one electron.

$\displaystyle 2PF_2I_{(l)}+2e^-\rightarrow P_2F_4_{(g)}$

$\displaystyle 2Hg \rightarrow Hg_2I_2_{(s)}+2e^{-}$

We can see that two electrons are tranferred. To identify the element being oxidized, we must find the element that is losing electrons. In this case, mercury is being oxidized.

An oxidation-reduction (redox) reaction is a reaction where electrons are transferred between two substances. When an atom is oxidized, it is called the reducing agent, and it loses a number of electrons. Similarly, a reduced atom is called the oxidizing agent, and gains the same number of electrons. A popular mnemonic to help remember is OIL RIG, or Oxidation is Loss of electrons, Reduction is Gain of electrons.

Iron (II) has a positive two charge: $\displaystyle Fe^{2+}$.

Phosphate has a negative three charge: $\displaystyle PO_4^{3-}$.

The initial compound must be constructed to cancel these charges. The dissociation is: $\displaystyle Fe_3(PO_4)_2\rightleftharpoons 3Fe^{2+}+2PO_4^{3-}$.

Calcium is in the second group of the periodic table, and is therefore going to have a $\displaystyle \small +2$ oxidation number. Hydroxide ions have a $\displaystyle \small -1$ charge. Calcium hydroxide will have the formula $\displaystyle \small Ca(OH)_2$.

Chloride ions have a $\displaystyle \small -1$ charge and hydrogen ions have a $\displaystyle \small +1$ charge. The formula for hydrochloric acid is $\displaystyle \small HCl$.

On the products side, water has the formula $\displaystyle \small H_2O$ and calcium chloride has the formula $\displaystyle \small CaCl_2$.

Now that we know all of the formulas, we can write our reaction:

$\displaystyle Ca(OH)_2+HCl\rightarrow H_2O+CaCl_2$

In order to balance the chloride atoms, we need to add coefficients.

$\displaystyle Ca(OH)_2+2HCl\rightarrow 2H_2O+CaCl_2$

The balanced reaction for the combustion of pentane is:

$\displaystyle C_{5}H_{12} + 8O_{2} \rightarrow 5CO_{2} + 6H_{2}O$

When balanced, oxygen gas has a coefficient of eight.

To balance the equation, it is easiest to leave oxygen and hydrogen for last. This means we should start with carbon.

$\displaystyle C_{5}H_{12} + O_{2} \rightarrow CO_{2} + H_{2}O$

$\displaystyle C_{5}H_{12} + O_{2} \rightarrow 5CO_{2} + H_{2}O$

Now that carbon is balanced, we can look at hydrogen.

$\displaystyle C_{5}H_{12} + O_{2} \rightarrow 5CO_{2} + 6H_{2}O$

Finally, we can balance the oxygen.

$\displaystyle C_{5}H_{12} + 8O_{2} \rightarrow 5CO_{2} + 6H_{2}O$

The final reaction uses five carbon atoms, twelve hydrogen atoms, and sixteen oxygen atoms per side.

The balanced chemical equation is:

$\displaystyle \tiny \small \small 2Al_(_s_) + Fe_2O_3_(_s_) \rightarrow Al_2O_3_(_s_) + 2Fe_(_s_)$

The mole ratio of iron oxide to solid iron is 1:2. You can set up the following proportion to solve:

$\displaystyle \frac{1mol\ Fe_2O_3}{2mol\ Fe}=\frac{10mol\ Fe_2O_3}{x}$

$\displaystyle x = (2)(10) = 20 mol\ Fe$

Potassium is a Group I element, so to get to a filled valence shell, it will lost one electron, yielding $\displaystyle K^+$.

Arsenic is a Group 5 element, so it needs to gain three electrons to obtain a filled valence shell, yielding $\displaystyle As^{-3}$.

In order to balance out the charges, the resultant salt will be $\displaystyle K_3As$.

First, we must know what ferrous sulfate is. Ferrous refers to $\displaystyle Fe^{2+}$ , and sulfate has the formula $\displaystyle SO_4^{-2}$. When we combine the two together we get $\displaystyle FeSO_4$.

Calcium is a divatent cation and iodide is a monovalent anion, so their salt is $\displaystyle CaI_2$. The ion exchange reaction is then:

$\displaystyle FeSO_4 + CaI_2 \rightarrow FeI_2 +CaSO_4$

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.

$\displaystyle Ba(NO_{3})_{2} (aq) + K_{2}SO_{4} (aq) \rightarrow 2KNO_{3} (aq) + BaSO_{4} (s)$

$\displaystyle Ba^{2+}+2NO_3^-+2K^++SO_4^{2-}\rightarrow2K^++2NO_3^-+BaSO_4(s)$

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.

$\displaystyle Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4(s)$

This is our net ionic equation.