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
anor277
Nov 17, 2016

#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?

Related questions
See all questions in Balancing Redox Equations Using the Oxidation Number Method
Impact of this question
8732 views around the world
You can reuse this answer
Creative Commons License