I have the following question stated in the science teacher exam:

Write the half-reaction equations for the following reaction by using oxidation numbers: $$\ce{Na2SO3 + Na2CrO4 -> Na2SO4 + Cr(OH)3 }$$

I am having some difficulty determining the half-reaction equations because the question doesn’t indicate either to the states of the species in the reaction nor to the medium of the reaction.

By determining oxidation numbers it is possible to see what is being oxidised and what is being reduced. I have indicated oxidation numbers below in parentheses:

$$ \ce{ \overset{\color{black}{(+1)}}{Na2} \overset{\color{red}{(+4)}}{S} \overset{\color{black}{(-2)}}{O_3} + \overset{\color{black}{(+1)}}{Na2}\overset{\color{green}{(+6)}}{Cr}\overset{\color{black}{(-2)}}{O_4} -> \overset{\color{black}{(+1)}}{Na2} \overset{\color{red}{(+6)}}{S} \overset{\color{black}{(-2)}}{O4} + \overset{\color{green}{(+3)}}{Cr}(\overset{\color{black}{(-2)}}{\ce{O}}\overset{\color{black}{(+1)}}{\ce{H}})3} $$

By supposing the reaction takes place in an acidic medium, I failed to answer the question.

But, by supposing the reaction takes place in the basic medium, I conclude to following two half-reactions:

Oxidation half-reaction: $$\ce{2OH- + Na2SO3 -> Na2SO4 + H2O + 2e-}$$ Reduction half- reaction: $$\ce{3e- + 4H2O + Na2CrO4 -> Cr(OH)3 +2Na+ + 5OH-}$$ The completely balanced equation: $$\ce{5H2O + 3Na2SO3 +2Na2CrO4 -> 3Na2SO4 + 2Cr(OH)3 + 4NaOH}$$

I am not sure about this solution. Are there, and where if there are, mistakes in the solution? What is the correct solution?

  • 2
    $\begingroup$ There's nothing "extraordinary" here and I don't see why you think there's something wrong. (Nice job with formatting though.) $\endgroup$ – Mithoron Apr 5 '19 at 0:43
  • $\begingroup$ Mithoron Thanks for the interest, what about the states of the species in the reaction? can a reaction occur in an acidic medium? $\endgroup$ – Adnan AL-Amleh Apr 5 '19 at 0:55
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    $\begingroup$ This reaction can't be in acidic medium because the chromium (III) hydroxide would dissolve. $\endgroup$ – Karsten Theis Apr 5 '19 at 3:04
  • $\begingroup$ @KarstenTheisHow the reaction proceeds? $\endgroup$ – Adnan AL-Amleh Apr 5 '19 at 5:45

You have rightfully assumed the reaction takes place in either basic or neutral medium, otherwise $\ce{Cr(OH)3}$ wouldn't be the product. This automatically suggests $\ce{H+}$ are not used for balancing half reactions. Your solution is fine for the provided reaction given that chromium(III) hydroxide stays precipitated. Here I've basically formatted your solution in a form I'm more familiar with:

\begin{align} \ce{\overset{+6}{Cr}O4^2- + 4 H2O + 3 e- &→ \overset{+3}{Cr}(OH)3 + 5 OH-} &|\cdot 2 \tag{red1}\\ \ce{\overset{+4}{S}O3^2- + 2 OH- &→ \overset{+6}{S}O4^2- + H2O + 2 e-} &|\cdot 3 \tag{ox1}\\ \hline \ce{2 CrO4^2- + 3 SO3^2- + 5 H2O &→ 2 Cr(OH)3 + 3 SO4^2- + 4 OH-} \tag{redox1} \end{align}

Resulting full redox equation:

$$\ce{2 Na2CrO4 + 3 Na2SO3 + 5 H2O → 2 Cr(OH)3 + 3 Na2SO4 + 4 NaOH} \tag{R1}$$

This implies that reaction begins in neutral medium; however there is remaining sodium hydroxide ($\ce{4 NaOH}$ on the right), so the pH will increase over time. This will result in further dissolving of $\ce{Cr(OH)3}$ in $\ce{NaOH}$ yielding $\ce{Na3[Cr(OH)6]}$ complex, which is likely the final product. Alternatively, addition of a base to this mix would also accelerate formation of chromium(III) hydroxocomplex. With this in mind, I'd say the more appropriate way to write the reaction is

\begin{align} \ce{\overset{+6}{Cr}O4^2- + 4 H2O + 3 e- &→ [\overset{+3}{Cr}(OH)6]^3- + 2 OH-} &|\cdot 2 \tag{red2}\\ \ce{\overset{+4}{S}O3^2- + 2 OH- &→ \overset{+6}{S}O4^2- + H2O + 2 e-} &|\cdot 3 \tag{ox2}\\ \hline \ce{2 CrO4^2- + 3 SO3^2- + 2 OH- + 5 H2O &→ 2 [Cr(OH)6]^3- + 3 SO4^2-} \tag{redox2} \end{align}

Resulting full redox equation:

$$\ce{2 Na2CrO4 + 3 Na2SO3 + 2 NaOH + 5 H2O → 2 Na3[Cr(OH)6] + 3Na2SO4} \tag{R2}$$

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