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?
$$\begin{array}{ccc}\ce{OH-} + (5) \xrightarrow{} & \ce{CrO4^2- + H+} & \\ & \ce{v} (6) & \\ & \ce{BaCrO4 (s)} & \ce{<=>[(7)][(8)]} & \ce{Ba^2+ + Cr2O7^2- + H2O}\\ & \text{(yellow precipitate)} & & \ce{v SO2 (g), H+} \\ & & & \ce{BaSO4(s)} + (9) + \ce{H2O} \end{array}$$
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?
Answer
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:
$$\ce{ CrO4^2- <=>[\ce{H+}][\ce{OH-}] Cr2O7^2- }$$
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:
$$\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}$$
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