Write the balanced net ionic equation for the following chemical reaction: Basic solutions of sodium sulfite, NaSO3, and potassium permanganate, KMnO4, are mixed.
To write the balanced net ionic equation for the given chemical reaction, let's break it down into two half-reactions: the reduction half-reaction and the oxidation half-reaction.
First, let's write the reduction half-reaction. In basic solution, sodium sulfite, NaSO3, can be reduced to sodium hydroxide, NaOH. The sulfite ion, SO3^2-, gains two electrons to become hydroxide ions, OH^-. Therefore, the reduction half-reaction is:
SO3^2- + 2H2O + 2e^- → 2OH^- (Equation 1)
Next, let's write the oxidation half-reaction. In basic solution, potassium permanganate, KMnO4, can be oxidized to form manganese dioxide, MnO2. The permanganate ion, MnO4^-, loses four electrons to become the dioxide ion, MnO2. Therefore, the oxidation half-reaction is:
MnO4^- → MnO2 + 4e^- (Equation 2)
Now, let's balance the number of electrons in both half-reactions. Multiply Equation 1 by 4 and Equation 2 by 1 to equalize the number of transferred electrons:
4(SO3^2- + 2H2O + 2e^-) → 4(2OH^-)
4MnO4^- → 4MnO2 + 16e^-
Finally, add the two half-reactions together and cancel out common species on both sides of the equation:
4(SO3^2- + 2H2O) + 4MnO4^- → 4(2OH^-) + 4MnO2
4SO3^2- + 8H2O + 4MnO4^- → 8OH^- + 4MnO2
Therefore, the balanced net ionic equation for the reaction between sodium sulfite, NaSO3, and potassium permanganate, KMnO4, in basic solution is:
4SO3^2- + 8H2O + 4MnO4^- → 8OH^- + 4MnO2