How do these cascade reactions work? A. $\ce{Cr2O7^2-}$  B. $\ce{H+}$  C. $\ce{OH-}$  D.
  $\ce{Cr^3+}$  E. $\ce{Ba^2+}$ 
  
  Which of the above is the appropriate ion for each blank in the
  following series of reactions?
  
  $$\begin{array}{ccc}\ce{OH-} + (5) \xrightarrow{} &amp; \ce{CrO4^2- + H+}
 &amp; \\  &amp; \ce{v} (6) &amp; \\  &amp; \ce{BaCrO4 (s)} &amp; \ce{&lt;=&gt;[(7)][(8)]} &amp;
 \ce{Ba^2+ + Cr2O7^2- + H2O}\\  &amp; \text{(yellow precipitate)} &amp; &amp; \ce{v
 SO2 (g), H+} \\  &amp; &amp; &amp; \ce{BaSO4(s)} + (9) + \ce{H2O} \end{array}$$


This question can also be viewed at https://i.stack.imgur.com/i08sR.jpg

So while studying, I came across this reactions' question, which I believe is called a 'cascade reaction'. I have no idea how to do it. I can figure out a few parts, like ($6$) is obviously $\text{E}$, but how do you do the others?
how do you do the others?


You do it like a detective. Even the slightest details matter, and you've to link a lot of concepts together.

The main theme you're missing in your question is this equilibrium:

$$\ce{ CrO4^2- &lt;=&gt;[\ce{H+}][\ce{OH-}] Cr2O7^2- }$$

Dichromates are supposed to be stable in an acidic medium. So using this you can figure out that ($7$) is $\ce{H+}$ and ($8$) is $\ce{OH-}$. Again for ($5$) you're adding a base which is converting it to a chromate. So ($5$) should be a dichromate $\ce{Cr2O7^2-}$.

So we're only left with ($9$) and $\ce{Cr^3+}$ and they are bound to go with each other. Since this is not an exam here's a reasoning too,

$\ce{K2Cr2O7}$ is a strong oxidizer in acidic mediums. If it's going to oxidize others, it will itself get reduced. To be precise here's the reaction that's taking place:

$$\ce{Cr2O7^2- + 14H+ + 6e- -&gt; 2Cr^3+ + 7H2O}$$