# Question 319b1

Mar 7, 2016

Here's how you could do that.

#### Explanation:

In order to write a balanced chemical equation for this redox reaction you must have an unbalanced chemical equation to work with.

An unbalanced chemical equation is usually provided to you, but since you didn't add it here, I'll take a few moments to describe what's happening here.

The permanganate anion, ${\text{MnO}}_{4}^{-}$, which contains manganese in its $\textcolor{b l u e}{+ 7}$ oxidation state, is an excellent oxidizing agent in an acidic medium, where it gets reduced to manganese(II) cations, ${\text{Mn}}^{2 +}$.

However, in a basic medium, like the one you have here, its oxidizing strength decreases significantly. In this case, the permanganate anion gets reduced to the manganate anion, ${\text{MnO}}_{4}^{2 -}$, in which manganese has an oxidation state of $\textcolor{b l u e}{+ 6}$.

The permanganese anion will oxidize sodium bisulfite, ${\text{NaHSO}}_{3}$, to sodium bisulfate, ${\text{NaHSO}}_{4}$. The reaction will also produce aqueous sodium manganate, ${\text{Na"_2"MnO}}_{4}$, potassium manganate, ${\text{K"_2"SO}}_{4}$, and water.

For simplicity, I"ll use the unbalanced net ionic equation as the starting point here. The equation includes the oxidation states for manganese and sulfur

${\text{H"stackrel(color(blue)(+4))("S")"O"_text(3(aq])^(-) + "OH"_text((aq])^(-) + stackrel(color(blue)(+7))("Mn")"O"_text(4(aq])^(-) -> stackrel(color(blue)(+6))("Mn")"O"_text(4(aq])^(2-) + "H"stackrel(color(blue)(+6))("S")"O"_text(4(aq])^(-) + "H"_2"O}}_{\textrm{\left(l\right]}}$

Write and balance the oxidation and reduction half-reactions - I wont add the states for simplicity

• the oxidation half-reaction

stackrel(color(blue)(+7))("Mn")"O"_4^(-) +1"e"^(-) -> stackrel(color(blue)(+6))("Mn")"O"_4^(2-)

Nothing to add here, the half-reaction is balanced with respect to mass and charge.

• the reduction half-reaction

${\text{H"stackrel(color(blue)(+4))("S")"O"_3^(-) -> "H"stackrel(color(blue)(+6))("S")"O"_4^(-) + 2"e}}^{-}$

Notice that the oxygen atoms are unbalanced. In order to balance them, use water molecules, $\text{H"_2"O}$, on the side that needs oxygen and protons, ${\text{H}}^{+}$, on the side that needs hydrogen.

This is the same method you'd use for reactions that take part in an acidic medium.

In this case, you will have

${\text{H"_2"O" + "H"stackrel(color(blue)(+4))("S")"O"_3^(-) -> "H"stackrel(color(blue)(+6))("S")"O"_4^(-) + 2"e"^(-) + 2"H}}^{+}$

Now, because you're in a basic solution, you have to neutralize the protons by adding hydroixde anions, ${\text{OH}}^{-}$, to both sides of the equation.

This will result in the formation of water molecules on the side that has protons.

2"OH"^(-) + "H"_2"O" + "H"stackrel(color(blue)(+4))("S")"O"_3^(-) -> "H"stackrel(color(blue)(+6))("S")"O"_4^(-) + 2"e"^(-) + overbrace(2"H"^(+) + 2"OH"^(-))^(color(purple)(2"H"_2"O"))

The balanced reduction half-reaction will thus be

$2 \text{OH"^(-) + color(red)(cancel(color(black)("H"_2"O"))) + "H"stackrel(color(blue)(+4))("S")"O"_3^(-) -> "H"stackrel(color(blue)(+6))("S")"O"_4^(-) + 2"e"^(-) + color(red)(cancel(color(black)(2)))"H"_2"O}$

$2 \text{OH"^(-) + "H"stackrel(color(blue)(+4))("S")"O"_3^(-) -> "H"stackrel(color(blue)(+6))("S")"O"_4^(-) + 2"e"^(-) + "H"_2"O}$

In any redox reaction, the number of electrons lost in oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

To balance the number of electrons transferred, multiply the oxidation half-reaction by $2$.

Add the two half-reactions to get

[stackrel(color(blue)(+7))("Mn")"O"_4^(-) +1"e"^(-) -> stackrel(color(blue)(+6))("Mn")"O"_4^(2-)] xx 2#
$2 \text{OH"^(-) + "H"stackrel(color(blue)(+4))("S")"O"_3^(-) -> "H"stackrel(color(blue)(+6))("S")"O"_4^(-) + 2"e"^(-) + "H"_2"O}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}$

$2 \text{OH"^(-) + 2"MnO"_4^(-) + "HSO"_3^(-) + color(red)(cancel(color(black)(2"e"^(-)))) -> 2"MnO"_4^(2-) + "HSO"_4^(-) + color(red)(cancel(color(black)(2"e"^(-)))) + "H"_2"O}$

The balanced net ionic equation will thus be

$2 \text{OH"^(-) + 2"MnO"_4^(-) + "HSO"_3^(-) -> "MnO"_4^(2-) + "MnO"_4^(2-) + "HSO"_4^(-) + "H"_2"O}$

Add the spectator ions and the states to get the balanced molecular equation

$2 {\text{KMnO"_text(4(aq]) + 2"NaOH"_text((aq]) + "NaHSO"_text(3(aq]) -> "NaHSO"_text(4(aq]) + "Na"_2"MnO"_text(4(aq]) + "K"_2"MnO"_text(4(aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$