Can you represent the oxidation of sulfite ion, #SO_3^(2-)#, to give sulfate ion, #SO_4^(2-)#, accompanied by the reduction of #Cu^(2+)# to copper metal?

Answer 1

#SO_3^(2-) + Cu^(2+)+ 2HO^(-) rarr SO_4^(2-) +Cu(s) + H_2O(l)#

It is usually best to balance the equations in acid and then adjust the final equations when basic conditions are specified.

#"Sulfur (IV+)"# #rarr# #"Sulfur(VI+)":#

#SO_3^(2-) + H_2Orarr SO_4^(2-) +2H^(+) + 2e^(-) #

For basic conditions, add #2xxHO^-# to EACH side #(HO^(-) + H^(+)rarrH_2O)#:

#SO_3^(2-) + 2HO^(-) rarr SO_4^(2-) +H_2O + 2e^(-) #

#"Copper (II+)"# #rarr# #"Copper (0)":#

#Cu^(2+) + 2e^(-) rarr Cu#

You recently posed another question of this kind: do you use the redox equation method?

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Answer 2

Caleb Adams

The oxidation of sulfite ion (SO3^2-) to sulfate ion (SO4^2-) accompanied by the reduction of Cu^2+ to copper metal can be represented by the following equation:

2 Cu^2+ + SO3^2- + 2 H2O → 2 Cu + SO4^2- + 4 H+

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.