# Why is this redox reaction not balanced? Please balance it. "H"_2"SO"_4(aq) + "K"_2"Cr"_2"O"_7(aq) + "SO"_2(g) -> "Cr"_2"SO"_4(aq) + "H"_2"O"(l) + "K"_2"SO"(aq)

May 29, 2017

$\text{Sulfur dioxide}$, ${\stackrel{+ I V}{\text{SO}}}_{2}$, is oxidized to sulfate, ${\stackrel{V I +}{\text{SO}}}_{4}^{- 2}$.

#### Explanation:

$S {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) \rightarrow S {O}_{4}^{2 -} + 4 {H}^{+} + 2 {e}^{-}$ $\left(i\right)$

And dichromate, ${\stackrel{+ V I}{\text{Cr}}}_{2} {O}_{7}^{2 -}$ is reduced to $C {r}^{3 +}$.

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O \left(l\right)$ $\left(i i\right)$

And these are both balanced with respect to mass and charge, as they must be...........and so we takes $3 \times \left(i\right) + \left(i i\right)$ to eliminate the electrons.......

$3 S {O}_{2} \left(g\right) + \cancel{6 {H}_{2} O \left(l\right)} + C {r}_{2} {O}_{7}^{2 -} + 2 \cancel{14} {H}^{+} + \cancel{6 {e}^{-}} \rightarrow 3 S {O}_{4}^{2 -} + \cancel{12 {H}^{+}} + \cancel{6 {e}^{-}} + 2 C {r}^{3 +} + \cancel{7} {H}_{2} O \left(l\right)$

And cancel out the common reagents to give........

$C {r}_{2} {O}_{7}^{2 -} + 3 S {O}_{2} \left(g\right) + 2 {H}^{+} \rightarrow 2 C {r}^{3 +} + 3 S {O}_{4}^{2 -} + {H}_{2} O \left(l\right)$

Which I think is balanced with respect to mass and charge, as indeed it must be if it is to reflect chemical reality.

All these redox reactions take a bit a practice. What you would observe in this reaction (and I hope that it will be demonstrated in a lab) is the orange-red colour of dichromate dissipate to give green $C {r}^{3 +}$ ion.

See here for other examples.

May 29, 2017

Your reaction has a typo or wherever you found it had an error. It should be:

${\text{H"_2"SO"_4(aq) + "K"_2"Cr"_2"O"_7(aq) + 3"SO"_2(g) -> "Cr"_2("SO"_4)_3(aq) + "H"_2"O"(l) + "K"_2"SO}}_{4} \left(a q\right)$

The number of oxygen atoms weren't balanced in your Google search. Trust but verify!

First off, for simplicity, let's examine the reaction of dichromate and sulfur dioxide, and ignore the potassium and sulfate counterions (but not the sulfate product). We can add them back later.

Now, some key words of note:

• "acidified" implies the redox reaction occurs in acidic solution (aqueous, because all ions dissolve in water), which means you can freely add ${\text{H}}^{+}$ in the process of balancing hydrogen atoms.
• "potassium dichromate oxidizing . . . " implies that the dichromate is the oxidizing agent, and hence is itself reduced. That gives you a hint to write the reduction half-reaction.
• " . . . sulfur dioxide to the sulfate ion" says that sulfate is the oxidation product. That gives you a hint to write the oxidation half-reaction.

From this, we can see how the following steps make sense:

1. Extract half-reactions involving each key ion and its oxidation or reduction to its respective product. (Include only the reactant and product for now, without balancing.)
2. Balance non-oxygen and non-hydrogen atoms using stoichiometric coefficients.
3. Balance oxygen atoms using $\boldsymbol{\text{H"_2"O}}$, since we are in aqueous solution.
4. Balance hydrogen atoms using $\boldsymbol{{\text{H}}^{+}}$, since we are in acidic, aqueous solution.
5. Balance the remaining charge by adding ${\text{e}}^{-}$, electrons, to the more positive side of the reaction (thus reducing the charge to balance with the other side) as an accounting scheme.
6. Add the two half-reactions together, scaling as necessary to cancel out the electrons.

You would find this useful, for example, if you have to re-use the same half-reaction multiple times. An example at my university is doing a series of reactions of permanganate with each of the halides.

WRITING THE REDUCTION HALF-REACTION

We learned that dichromate was reduced (the oxidation state became less positive, or the number of oxygen atoms decreased), so... the starting half-reaction could be this:

${\text{Cr"_2"O"_7^(2-)(aq) -> "Cr}}^{3 +} \left(a q\right)$

Of course, chromium could have been reduced to ${\text{Cr}}^{2 +}$ instead; we don't really know for sure, and this is one possible result that is based on your googled example. This diagram gives ${\text{Cr}}^{2 +}$ and ${\text{Cr}}^{0}$ as other possibilities in acidic solution through experiencing a negative ${E}_{\text{cell}}$. We'll go with this result, because it's the process that matters. Now, follow the above steps.

Balance the non-oxygen, non-hydrogen atoms.

${\text{Cr"_2"O"_7^(2-)(aq) -> color(red)(2)"Cr}}^{3 +} \left(a q\right)$

Now, the oxygen atoms.

"Cr"_2"O"_7^(2-)(aq) -> color(red)(2)"Cr"^(3+)(aq) + color(red)(7"H"_2"O"(l))

Now, the hydrogen atoms that the water molecules added just skewed. Remember, we can add ${\text{H}}^{+}$ since the solution is acidified.

$\textcolor{red}{14 \text{H"^(+)(aq)) + "Cr"_2"O"_7^(2-)(aq) -> color(red)(2)"Cr"^(3+)(aq) + color(red)(7"H"_2"O} \left(l\right)}$

All the atoms are balanced, and so we look at the charge.

$14 \left({1}^{+}\right) + \left({2}^{-}\right) \text{ vs. } 2 \left({3}^{+}\right) + 0$

$\implies + 12 \text{ vs. } + 6$

$\implies$ add $6 {e}^{-}$ to the left side.

$\textcolor{g r e e n}{6 {e}^{-} + 14 \stackrel{\textcolor{b l u e}{+ 1}}{\text{H"^(+))(aq) + stackrel(color(blue)(+6))("Cr"_2)stackrel(color(blue)(-2))("O"_7^(2-))(aq) -> 2stackrel(color(blue)(+3))("Cr"^(3+))(aq) + 7stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)}$

There, now this is a balanced reduction half-reaction! If you notice, I've also added in each atom's oxidation state at the end. We see that $\stackrel{+ 6}{\text{Cr") -> stackrel(+3)("Cr}}$.

We now move on to the oxidation half-reaction.

WRITING THE OXIDATION HALF-REACTION

Same process, but now we write this to begin with:

${\text{SO"_2(g) -> "SO}}_{4}^{2 -} \left(a q\right)$

Go through the same steps as above to obtain:

$\textcolor{g r e e n}{2 \stackrel{\textcolor{b l u e}{+ 1}}{{\text{H"_2)stackrel(color(blue)(-2))("O")(l) + stackrel(color(blue)(+4))("S")stackrel(color(blue)(-2))("O"_2)(g) -> stackrel(color(blue)(+6))("S")stackrel(color(blue)(-2))("O"_4^(2-))(aq) + 4stackrel(color(blue)(+1))("H}}^{+}} \left(a q\right) + 2 {e}^{-}}$

We see that $\stackrel{+ 4}{\text{S") -> stackrel(+6)("S}}$.

OVERALL REACTION

Now, we add the two half-reactions together, trying to cancel out the electrons. If we multiply the oxidation half-reaction by $3$, that works out.

$\textcolor{g r e e n}{\cancel{6 {e}^{-}} + 14 \stackrel{\textcolor{b l u e}{+ 1}}{\text{H"^(+))(aq) + stackrel(color(blue)(+6))("Cr"_2)stackrel(color(blue)(-2))("O"_7^(2-))(aq) -> 2stackrel(color(blue)(+3))("Cr"^(3+))(aq) + 7stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)}$
$3 \left(\textcolor{g r e e n}{2 \stackrel{\textcolor{b l u e}{+ 1}}{{\text{H"_2)stackrel(color(blue)(-2))("O")(l) + stackrel(color(blue)(+4))("S")stackrel(color(blue)(-2))("O"_2)(g) -> stackrel(color(blue)(+6))("S")stackrel(color(blue)(-2))("O"_4^(2-))(aq) + 4stackrel(color(blue)(+1))("H}}^{+}} \left(a q\right) + \cancel{2 {e}^{-}}}\right)$
$\text{-------------------------------------------------------------------------------}$
$14 {\text{H"^(+)(aq) + "Cr"_2"O"_7^(2-)(aq) + 6"H"_2"O"(l) + 3"SO"_2(g) -> 2"Cr"^(3+)(aq) + 7"H"_2"O"(l) + 3"SO"_4^(2-)(aq) + 12"H}}^{+} \left(a q\right)$

Now cancel out matching substances on each side.

stackrel(2)cancel(14)"H"^(+)(aq) + "Cr"_2"O"_7^(2-)(aq) + cancel(6"H"_2"O"(l)) + 3"SO"_2(g) -> 2"Cr"^(3+)(aq) + cancel(7)"H"_2"O"(l) + 3"SO"_4^(2-)(aq) + cancel(12"H"^(+)(aq))

$\implies \textcolor{b l u e}{2 {\text{H"^(+)(aq) + "Cr"_2"O"_7^(2-)(aq) + 3"SO"_2(g) -> 2"Cr"^(3+)(aq) + "H"_2"O"(l) + 3"SO}}_{4}^{2 -} \left(a q\right)}$

In most cases it's OK to leave this as it is. But since you wanted the counterions in there, let's put them back in there.

PUTTING BACK THE COUNTERIONS

There's an error in your reaction you've given. The oxygen atoms are NOT balanced! Here's how to proceed from what we just got up above, which is balanced.

• Add in one sulfate counterion on the $2 {\text{H}}^{+}$ on the left-hand side, and consequently one on the right-hand side.
• Combine $2 {\text{Cr}}^{3 +}$ with $3 {\text{SO}}_{4}^{2 -}$ for dat charge balance. Also, this compound is a solid when anhydrous, but addition of a sufficient reducing agent like ${\text{SO}}_{2} \left(g\right)$ allows for the compound to be aqueous.
• Add in the potassium counterions (two on each side).

So, it should be:

$\textcolor{b l u e}{{\text{H"_2"SO"_4(aq) + "K"_2"Cr"_2"O"_7(aq) + 3"SO"_2(g) -> "Cr"_2("SO"_4)_3(aq) + "H"_2"O"(l) + "K"_2"SO}}_{4} \left(a q\right)}$

There you have it, the reaction is now balanced, with all the counterions as well!