Heptane: C₇H₁₆. Combustion is when a molecule reacts with O₂ and the products are CO₂ and H₂O. Balancing gives 7 CO₂, 8 H₂O, 1 heptane, and 11 O₂

The longer the hydrocarbon chain, the greater the amount of combustion products (CO2 and H20) generated. Branched molecules such as isopropane and isobutane are more difficult to combust than their straight-chain counterparts.

An exothermic reaction will have a negative \(\displaystyle \Delta H\) value, indicating that it releases heat. Conversely, an endothermic reaction will have a positive \(\displaystyle \Delta H\) value, indicating a consumption of heat.

A combustion reaction releases heat; thus it must have a negative \(\displaystyle \Delta H\) value and be exothermic.

Exergonic reactions have a negative \(\displaystyle \Delta G\) value, indicating spontaneity, while endergonic reactions are non-spontaneous. While most combustion reactions will be non-spontaneous, it is impossible to draw this conclusion for certain without knowing more about the reaction. The only thing we know for certain is that heat is released, and the reaction is exothermic.

When you balance the redox reaction in acidic conditons, there are 6Fe²⁺, 1 Cr₂O₇^2–, 14 H⁺, 6 Fe³⁺, 2 Cr³⁺, and 7 H₂O. Don't forget to add the 1 in front of the Cr₂O₇^2–

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.

The common factor between 1 e^- and 5 e^- is 5.  Therefore the number of electrons involved is 5 e^-.

The common factor between 2 e^- and 5 e^- is 10.  Therefore the number of electrons involved is 10 e^-.

Add them together:

\(\displaystyle 3 H_2 O + Br^- + 2 MnO_4^- + 8 H^+ + 6 e^- \rightarrow 2 MnO_2 + BrO_3^- + 6 H^+ + 4 H_2 O + 6 e^-\)

Simplify: 

\(\displaystyle Br^- + 2 MnO_4^- + 2 H^+ \rightarrow 2 MnO_2 + BrO_3^- + H_2 O\)

Add Hydroxides to each side to counter H⁺_.

\(\displaystyle Br^- + 2 MnO_4^- + 2 H_2 O \rightarrow 2 MnO_2 + BrO_3^- + H_2 O+2 OH^-\)

Simplify:

\(\displaystyle Br^- + 2 MnO_4^- + H_2 O \rightarrow 2 MnO_2 + BrO_3^- +2 OH^-\)

Add the equations together

\(\displaystyle 2Sb+4H_2 O + 3 O_2 + 6 H^+ + 6e^- \rightarrow 2SbO_2^-+8H^++3 H_2 O_2 + 6e^-\)

Simplify

\(\displaystyle 2Sb+4H_2 O+3 O_2 \rightarrow 2SbO_2^-+2H^++3H_2 O_2\)

Add 2 OH^- to each side to cancel out the H⁺_.

\(\displaystyle 2Sb+4H_2 O+3 O_2 + 2 OH^- \rightarrow 2SbO_2^-+2H_2 O+3H_2 O_2\)

Simplify:

\(\displaystyle 2Sb+2 H_2 O+3 O_2 + 2 OH^- \rightarrow 2SbO_2^-+3H_2 O_2\)

Recall that oxidation always occurs at the anode (in both the electrochemical and galvanic cells). \(\displaystyle Fe\) loses two electrons in this case to become \(\displaystyle Fe^{2+}\). The presence of \(\displaystyle Fe^{2+}\) is hinted by the ionic compound \(\displaystyle FeCl_2\)_.