Cu(s)+2H2SO4(ag)----> CuSO4(ag)+2H20+SO2 (g)
This is a big equation where do I start and how do i get the end result which is "Some of the S atoms in the H2SO4 are reduced to for SO2. (+6 to +4)
2007-03-03 14:53:57 · 2 answers · asked by Toothie 2 in Science & Mathematics ➔ Chemistry
In this equation, Cu is oxidized. H2SO4 is the oxidizing agent. Hence H2SO4 is itself reduced. The oxidation state of S in H2SO4 is +6 and in SO2 is +4. A decrease in oxidation state refers to reduction.
2007-03-03 15:15:05 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Well the best way to think about oxidation/ reduction reactions, is the reduced gains an electron(s) and oxidized loses electron(s). So based upon your statement of (+6 to +4) it would be reduced, but this might give you a place to start looking up the answer to your question.
2007-03-03 15:05:20 · answer #2 · answered by brian_holinsworth1 2 · 0⤊ 0⤋