Balancing a reaction where dichromate ion reacts with ethanol to give Cr(III) and carbon dioxide
Let's get the simple question out of the way first:

How do you balance a redox reaction lacking a molecular reactant on the product side? I am referring to the atom or molecule that changes oxidation state when I say molecular reactant.

You can not balance it then. Also, you wouldn't be given such a reaction and is always a typo when they've said that it is a redox reaction.

Now, onto the bit more complicated question.

Another matter of concern is the answer in the answer key; the answer in acidic solution is, apparently:
$$\require{cancel}\ce{2Cr2O7^2- + 16H+ + C2H5OH -&gt; 4Cr^3+ + 2CO2 + 11H2O}$$
In a basic solution:
$$\ce{2Cr2O7^2- + 5H2O + C2H5OH -&gt; 4Cr^3+ + 2CO2 + 16OH-}$$
[sic] It seems like in the answer for the reaction in acidic medium, the left side of the reaction has a charge of +14 and the right side has a charge of +12.
However,  in the basic medium the answer has the left side with a total charge of -2 and the right side -4. How can these answers be correct if the charge isn't balanced?

A simple explanation would be that the excess charges that you feel are there are actually not. Try breaking the given reaction in the answer into each atom with their own oxidation state. I'll do the one for acidic medium, and the other will be up to the reader.
$$\ce{2Cr2O7^2- + 16H+ + C2H5OH -&gt; 4Cr^3+ + 2CO2 + 11H2O}$$
Here, we start from the reactant side, we assume the following for simplicity:

$\ce{C-H}$ bond leads to positive charge on $\ce{H}$ and a negative charge on $\ce{C}$
$\ce{Cr-O}$ bond leads to positive charge on $\ce{Cr}$ and a negative charge on $\ce{O}$
$\ce{C-O}$ bond leads to positive charge on $\ce{C}$ and a negative charge on $\ce{O}$

\begin{array}{lc} \hline
&amp;\text{Reactants} &amp;\text{Products} \\ \hline
\ce{Cr} &amp; \ce{Cr^6+} &amp; \ce{Cr^3+} \\
\ce{O} &amp; \ce{O^2-} &amp; \ce{O^2-} \\
\ce{H} &amp; \ce{H+} &amp; \ce{H+} \\
\ce{C} &amp; \ce{C-,C^3-}  &amp; \ce{C^4+} \\ \hline
\end{array}
Now, we split the actual equation into this form:
$$\ce{4Cr^6+ + \cancel{22\mathrm H^+} + C- + C^3- +\cancel{15\mathrm O^{2-}} -&gt; 4Cr^3+ + 2C^4+ + \cancel{15\mathrm O^{2-}} + \cancel{22\mathrm H^+}}$$
Which simplifies the reaction to:
$$\ce{4Cr^6+ +C- +C^3- -&gt; 4Cr^3+ +2C^4+}$$
Adding the charges on both sides we get:
\begin{align}
\text{Charge on reactant}=(4\times 6 + (-1) + (-3)) = 20 \\
\text{Charge on products}=(4\times 3 + 2\times4)= 20
\end{align}
Therefore the charge on both sides are equal ...
Note: The reaction in basic medium uses $\ce{CrO4^2-}$ and not $\ce{Cr2O7^2-}$
$$\ce{4CrO4^2- + 7H2O + C2H5OH -&gt; 4Cr^3+ + 2CO2 + 20OH-}$$