Question

Given that the sulfate(IV) ion, `SO_2^(-2)`, is converted to the sulfate(VI) ion, `SO_4^(-2)`, in the presence of water, deduce the balanced equation for the redox reaction between `Cr_2O_7^(-2)` (aq) and `SO_3^(-2)`?

Answer 1

`Cr_2O_7^(2-)+3SO_3^(2-) +8H^(+)rarr 2Cr^(3+)+ 3SO_4^(2-) + 4H_2O`

Explanation:

`"Oxidation half equation:"`

`SO_3^(2-) +H_2O rarr SO_4^(2-) + 2H^(+) + 2e^-` `(i)`

`"Reduction half equation:"`

`Cr_2O_7^(2-) +14H^(+) + 6e^(-)rarr 2Cr^(3+) + 2H^(+) + 7H_2O` `(ii)`

Both equations are balanced with respect to mass and charge; as indeed they must be if they reflect chemical reality. So we simply cross-mulitply to eliminate the electrons, `3xx(i)+(ii):`

`Cr_2O_7^(2-)+3SO_3^(2-) +8H^(+)rarr 2Cr^(3+)+ 3SO_4^(2-) + 4H_2O`

Given that the sulfate and sulfite ions are colourless, what would be the macroscopic colour change observed in this reaction?