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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?

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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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    $\begingroup$ Thank you! The only question I have left is, how do you know that $\text{Cr}^{3+}$ goes in the spot $9$? I understand that you would have some Chromium, but why wouldn't the dichromate ion stay in place? $\endgroup$ – Equinox May 26 '17 at 20:26
  • $\begingroup$ @Neal If you look at the oxidation states, $\ce{SO2}$ changes into $\ce{SO4^2-}$, i.e. sulfur gets oxidized. So, dichromate ion must undergo reduction. Thus, chromium changes from a $+6$ oxidation state to $+3$. $\endgroup$ – Berry Holmes May 27 '17 at 0:44
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As noted by @Berry, hydroxide ion converts perchromate ion to chromate ion and a proton converts chromate ion to perchromate ion.

$$\ce{ CrO4^2- <=>[\ce{H+}][\ce{OH-}] Cr2O7^2- }$$

So, in this case you figured out 7 and 8 to be proton(B) and hydroxide ion(C) and thus 5 would be (A) from above equation.

Since you want barium chromate from chromate ion, you just add any barium salt(E).

And last one would obviously would be left out option i.e (D). Unfortunately, I couldn't find any reaction for this. Only thing I can think of is that sulfur dioxide is acting as reducing agent(reduce perchromate to chromium(III)(D)).


Full reactions

  1. $$\ce{K2Cr2O7(A) + 2KOH(C) → 2K2CrO4 + H2O}$$

If you use potassium ions as counter ions.

Potassium dichromate react with potassium hydroxide to produce potassium chromate and water. Potassium hydroxide - concentrated solution.(source)

  1. $$\ce{Ba(OH)2(E) + K2CrO4 → BaCrO4 + 2KOH}$$

Barium hydroxide react with potassium chromate to produce barium chromate and potassium hydroxide.(source)

  1. $$\ce{2BaCrO4 + 2HCl(B) → BaCr2O7 + BaCl2 + H2O}$$

Barium chromate react with hydrogen chloride to produce barium dichromate, barium chloride and water. Hydrogen chloride - diluted solution.(source).

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