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I have only started to learn balancing redox reactions, and a recent question has me confused.

The question is :

Balance in both acidic medium and basic medium: $$\ce{Cr2O7^2- + C2H5OH -> Cr^3+ + CO2}$$

I usually look for symmetries between the reactant and product sides of a reaction to pair reactants with the corresponding products to form the half reactions, as per the first steps of balancing a redox reaction. I'm having a difficult time doing that here; what am I supposed to pair Cr with? I would personally pair it with the $\ce{Cr2O7^2-}$ molecule but that's just by process of elimination because I would naturally pair $\ce{C2H5OH}$ with $\ce{CO2}$ due to the stronger resemblance they have with each other.

Another matter of concern is the answer in the answer key; the answer in acidic solution is, apparently:

$$\ce{2Cr2O7^2- + 16H+ + C2H5OH -> 4Cr^3+ + 2CO2 + 11H2O}$$

In a basic solution:

$$\ce{2Cr2O7^2- + 5H2O + C2H5OH -> 4Cr^3+ + 2CO2 + 16OH-}$$

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?

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.

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  • $\begingroup$ I see you put in a lot of effort into making those reactions look like actual reactions. There is a simpler way to do that using $\ce{}$ which is used for chemical compounds.. If you wish you can look at the actual code by hitting edit $\endgroup$ – Safdar Sep 20 at 14:31
  • $\begingroup$ For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details $\endgroup$ – Safdar Sep 20 at 14:31
  • $\begingroup$ Thank you, I saw that I could use HTML and just got blindsided and went with that $\endgroup$ – Matthew S. Sep 20 at 14:32
  • $\begingroup$ Also, I think that there was a typo which I may have accidentally corrected that may have removed part of your question. It was $\ce{Cr^3+}$ and not $\ce{Cn^3+}$ $\endgroup$ – Safdar Sep 20 at 14:33
  • $\begingroup$ You made a couple bad edits IMO. For one, the changed title is not the question I'm asking, I'm asking the original title question that I had because I assume that this is not the only reaction with the same problem. There were also other unnecessary pedantic edits that changed the meaning of my question in an adverse way; it was supposed to be Cn<sup>3+</sup> unless this the reaction at the bottom of the page on this link under the section titled "Practice Problems" was erroneous and contained the typo $\endgroup$ – Matthew S. Sep 20 at 14:40
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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: $$\require{cancel}\ce{2Cr2O7^2- + 16H+ + C2H5OH -> 4Cr^3+ + 2CO2 + 11H2O}$$ In a basic solution: $$\ce{2Cr2O7^2- + 5H2O + C2H5OH -> 4Cr^3+ + 2CO2 + 16OH-}$$ [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.

$$\ce{2Cr2O7^2- + 16H+ + C2H5OH -> 4Cr^3+ + 2CO2 + 11H2O}$$ 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 &\text{Reactants} &\text{Products} \\ \hline \ce{Cr} & \ce{Cr^6+} & \ce{Cr^3+} \\ \ce{O} & \ce{O^2-} & \ce{O^2-} \\ \ce{H} & \ce{H+} & \ce{H+} \\ \ce{C} & \ce{C-,C^3-} & \ce{C^4+} \\ \hline \end{array}

Now, we split the actual equation into this form: $$\ce{4Cr^6+ + \cancel{22\mathrm H^+} + C- + C^3- +\cancel{15\mathrm O^{2-}} -> 4Cr^3+ + 2C^4+ + \cancel{15\mathrm O^{2-}} + \cancel{22\mathrm H^+}}$$

Which simplifies the reaction to:

$$\ce{4Cr^6+ +C- +C^3- -> 4Cr^3+ +2C^4+}$$

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 $\ce{Cr2O7^2-}$

$$\ce{4CrO4^2- + 7H2O + C2H5OH -> 4Cr^3+ + 2CO2 + 20OH-}$$

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Dichromate (orange) is present in acidic medium while chromate (yellow) prevails in basic medium. Here are stepwise solutions in both media.
Acidic Medium:
Recognizing that chromium is the oxidant, balance the chromium atoms.
$$\ce{Cr2O7^2- -> 2Cr^3+}$$ Because the reaction is conducted in aqueous acid, only water and protons are available to balance the half reaction. Balance the oxygens using $\ce{H2O}$.
$$\ce{Cr2O7^2- -> 2Cr^3+ + 7H2O}$$ Now balance the protons. $$\ce{14H+ + Cr2O7^2- ->[{+6e}] 2Cr^3+ + 7H2O}\label{1}\tag{1}$$ The carbon in ethanol is being oxidized. $$\ce{C2H5OH -> 2CO2}$$ Balance the oxygen with water added to the left side. $$\ce{3H2O +C2H5OH -> 2CO2}$$ Add protons to the right side. $$\ce{3H2O +C2H5OH ->[{-12e}] 2CO2 + 12H+}\label{2}\tag{2}$$ Multiply equation $\eqref{1}$ by 2 to balance electrons in both half reactions. $$\ce{28H^+ + 2Cr2O7^2- ->[{+12e}] 4Cr^3+ + 14H2O}\label{3}\tag{3}$$ Add equations $\eqref{2}$ and $\eqref{3}$. \begin{align} \ce{28H^+ + 2Cr2&O7^2- + 3H2O +C2H5OH \\ &->[{+12e}] 4Cr^3+ + 14H2O + 2CO2 + 12H+}\label{4}\tag{4} \end{align} Simplify duplicates on either side of equation $\eqref{4}$. $$\ce{16H^+ + 2Cr2O7^2- +C2H5OH -> 4Cr^3+ + 11H2O + 2CO2}\label{5}\tag{5}$$ Equation $\eqref{5}$ is the balanced oxidation in acidic medium.
Basic Medium:
In basic medium the chromate anion is dominate. Only water and hydroxide are available to balance oxygen and hydrogen. $$\ce{CrO4^2- -> Cr^3+}$$ Every oxygen atom requires two hydroxides where they are needed and one water on the other side of the equation ($\ce{2OH - H2O = O}$). Hydrogens are balanced. $$\ce{4H2O +CrO4^2- ->[{+3e}]Cr^3+ + 8OH^-}\label{6}\tag{6}$$ Now deal with ethanol. $$\ce{C2H5OH->2CO2}$$ Balance oxygen. $$\ce{6OH^- + C2H5OH->2CO2 + 3H2O}\tag{7}$$ Balance hydrogens ($\ce{H2O - OH^- = H}$). Water on the right. Hydroxide on the left. $$\ce{12OH^- + C2H5OH->[{-12e}]2CO2 + 9H2O}\label{8}\tag{8}$$ Multiply equation $\eqref{6}$ by 4 to balance electrons. $$\ce{16H2O +4CrO4^2- ->[{+12e}]4Cr^3+ + 32OH^-}\label{9}\tag{9}$$ Add equations $\eqref{8}$ and $\eqref{9}$. \begin{align} \ce{16H2O +4Cr&O4^2- + 12OH^- + C2H5OH \\ &->4Cr^3+ + 32OH^- + 2CO2 + 9H2O} \label{10}\tag{10} \end{align} Simplify equation $\eqref{10}$.

$$\ce{4CrO4^2- + C2H5OH + 7H2O -> 4Cr^3+ + 2CO2 + 20OH^-}\label{11}\tag{11}$$ Equation $\eqref{11}$ is balanced in atoms and electrons.
Postscript: I suspect that this reaction, in acid medium at least, is not a known reaction. If it were, how could a Kuhn-Roth C-methyl determination possibly be accomplished? The process depends upon the quantification of acetic acid.

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  • $\begingroup$ You can use $\tag{}$ for tagging equations and a combination of $\label{<label-name>}$ and $\eqref{<label-name>}$ to reference equations. I suspect you know this, but I mentioned it just in case. $\endgroup$ – Safdar Sep 21 at 12:27
  • $\begingroup$ @Safdar: After 3 years on ChemSE, this was my first attempt at inline equations. I can learn from your edit. $\endgroup$ – user55119 Sep 21 at 13:27
  • $\begingroup$ @Safdar Close to 6 years here and I don't think I even heard that ^ can be done with MathJax, you really dig this stuff ;D $\endgroup$ – Mithoron Oct 8 at 22:53

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