Question

A sample of a chromium-containing alloy weighing 3.450 g was dissolved in acid, and all the chromium in the sample was oxidized to 2cro42–. it was then found that 3.18 g of na2so3 was required to reduce the 2cro42– to cro2– in a basic solution, with the so32– being oxidized to so42–. write a balanced equation for the reaction of 2cro42– with so32- in a basic solution.

Answer 1

The balanced equation for the reaction of chromate ion (CrO4^2-) with sulfite ion (SO3^2-) in a basic solution where the sulfite is oxidized to sulfate and the chromate is reduced to chromite is 3SO3^2-(aq) + 2CrO4^2-(aq) + 2OH^-(aq) → 3SO4^2-(aq) + 2CrO2^-(aq) + H2O(l).

Step-by-step explanation:

The question asks for a balanced chemical reaction between chromate ion (CrO42-) and sulfite ion (SO32-) in basic solution. To balance this redox reaction, we must consider both the oxidation and reduction half-reactions and ensure that the number of electrons lost in oxidation equals the number gained in reduction, also making sure to balance other elements and charges, particularly in a basic solution.

In the basic solution, hydroxide ions (OH-) will participate in the balancing process. The sulfite ion (SO32-) is oxidized to sulfate ion (SO42-), and the chromate ion (CrO42-) is reduced to chromite ion (CrO2-).

The balanced equation for the reaction is:

3SO32-(aq) + 2CrO42-(aq) + 2OH-(aq) → 3SO42-(aq) + 2CrO2-(aq) + H2O(l).

Answer 2

Reduction half reaction

2H_2O + CrO_4^2- + 3e -> CrO_2^- + 4OH^-

Oxidation half reaction

2OH^- + SO_3^2- -> SO_4^2- + H_2O + 2e

Balanced overall equation

H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-