# Balancing a Redox Equation with Half-Reactions

• RaamGeneral
In summary, the conversation discusses balancing an equation involving sulfur dioxide and potassium dichromate in an acidic solution. The use of half-reactions is suggested, and after considering the role of K+ as a spectator ion, the final balanced equation is determined to be 3SO2 + K2Cr2O7 + H2SO4 → K2SO4 + Cr2(SO4)3 + H2O.
RaamGeneral
It's not clear to me how I can balance the following equation, most of all because of K which I don't know how to deal with:

$$\mathrm{SO_2 + K_2Cr_2O_7 \to Cr_2(SO_4)_3 + SO_3}$$ [this is the text of the exercise]

It's acidic solution for sulfuric acid, so the equation can be better written so:

$$\mathrm{SO_2 + K_2Cr_2O_7 + H_2SO_4 \to Cr_2(SO_4)_3 + SO_3 + H_2O + K}$$ (I assume K has to be added, preserving the charge in both members)

To use the half-reaction method I write the ionic form:

$$\mathrm{SO_2+K^+ + Cr_2O_7^{2-}+H^+ \to Cr^{3+}+SO_3+H_2O + K}$$

Writing and balancing three half-reactions (can it be?) I get:

$$\mathrm{4SO_2 + K_2Cr_2O_7+3H_2SO_4 \to 4SO_3 + Cr_2(SO_4)_3+2K+3H_2O}$$Is it correct?

Thank you.

Hint: K+ is just a spectator, and SO3 in water means just SO42-.

Ok, your hint was foundamental to me.
I have already thought about K+ being spectator but I "forgot" that
$$\mathrm{ SO_3 + H_2O\to H_2SO_4\to 2H^+ + SO_4^{2-} }$$ (right?) so I didn't know what to do without K in the second member.$$\mathrm{SO_2 + K_2Cr_2O_7 + H_2SO_4\to Cr_2(SO_4)_3 + SO_3 + H_2O + 2K^+ }$$
$$\mathrm{SO_2 + K_2Cr_2O_7 \to Cr_2(SO_4)_3 + 2K^+ }$$
ionic form: $$\mathrm{SO_2 + Cr_2O_7^{2-}\to Cr^{3+} + SO_4^{2-} }$$

Balancing the two half-reactions:

$$\mathrm{3SO_2 + Cr_2O_7^{2-} + 2H^+\to 3SO_4^{2-} + 2Cr^{3+} + H_2O }$$

And finally:

$$\mathrm{3SO_2 + K_2Cr_2O_7 + H_2SO_4\to K_2SO_4 +Cr_2(SO_4)_3 + H_2O }$$I'm not so sure though. Is it correct?

Last edited:
Looks OK to me.

## 1. What is a redox equation?

A redox equation is a chemical equation that shows the transfer of electrons between reactants and products. It involves two half-reactions, one where electrons are lost (oxidation) and one where electrons are gained (reduction).

## 2. How do I balance a redox equation?

To balance a redox equation, you must first identify the oxidation numbers of each element in the reactants and products. Then, you can add coefficients to the reactants and products to ensure that the number of electrons lost in oxidation is equal to the number of electrons gained in reduction.

## 3. What is the purpose of balancing a redox equation?

The purpose of balancing a redox equation is to accurately represent the chemical reaction and ensure that the law of conservation of mass and charge is followed. A balanced equation also allows for the calculation of reaction stoichiometry.

## 4. How do I know if a redox equation is balanced?

A redox equation is balanced when the number of atoms of each element is the same on both sides of the equation, and the total charge is also equal. Additionally, the number of electrons lost in oxidation must be equal to the number of electrons gained in reduction.

## 5. Can a redox equation be balanced in acidic and basic solutions?

Yes, a redox equation can be balanced in both acidic and basic solutions. In acidic solutions, H+ ions and water molecules are added to balance the charges and atoms. In basic solutions, OH- ions and water molecules are added to balance the charges and atoms.

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