Question

S + 02
C + 02
Cu0 + H2SO4

Answer 1

Answer:

S + 02= S0 - 4 e- → SIV (oxidation); 2 O0 + 4 e- → 2 O-II (reduction)

C + 02= C

Cu0 + H2SO4= CuO + H2SO4 → CuSO4 + H2O