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)#?

1 Answer
Nov 17, 2016

Answer:

#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?