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
Step by step explanation is given above.
I don’t know if ur Answer is correct you have a lot of O. But the answer would be : They are shiny and bend without breaking.