No. Copper will not replace hydrogen in sulfuric acid because it is less reactive than hydrogen.
Cu is oxidized. The oxidation number goes from 0 in Cu to +2 in CuSO4. S is reduced. The oxidation number goes from +6 in H2SO4 to +4 in SO2. The oxidizing agent is H2SO4 since it causes Cu to be oxidized. The reducing agent is Cu since it causes S in H2SO4 to be reduced.
In order for something to act as an oxidizing agent, it itself must be reduced, or put in other terms, it must be able to gain electrons. And for the record, the H2SO4 in order to react with Cu and to oxidize it, must be HOT and CONCENTRATED. Just putting Cu in some H2SO4 will not produce a reaction. The reaction of Cu + H2SO4 is as follows: Cu(s) + 2H2SO4(l) ---> Cu2SO4(s) + SO2(g) + 2H2O(g) ... note the states of each. In this reaction, the S of H2SO4 has gained electrons to form SO2, so it has been reduced and it has oxidized the Cu which went from Cu(s) to Cu^2+ in CuSO4. HCl is not capable of gaining electrons from Cu so it cannot oxidize the Cu.
H, Mg, Zn, Cu
It is exothermic. Take for example H2SO4 H2SO4 -> H+ + HSO4- This is very exothermic
It is related to concentration, which you do not give.
CuSO4 Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O
Cu is oxidized. The oxidation number goes from 0 in Cu to +2 in CuSO4. S is reduced. The oxidation number goes from +6 in H2SO4 to +4 in SO2. The oxidizing agent is H2SO4 since it causes Cu to be oxidized. The reducing agent is Cu since it causes S in H2SO4 to be reduced.
The oxidation number of hydrogen (H) in H2SO4 is +1.
Cu + H2SO4 = CuSO4 + H2
The balanced equation for the reaction between copper(II) hydroxide (Cu(OH)2) and sulfuric acid (H2SO4) is Cu(OH)2 + H2SO4 → CuSO4 + 2H2O.
In order for something to act as an oxidizing agent, it itself must be reduced, or put in other terms, it must be able to gain electrons. And for the record, the H2SO4 in order to react with Cu and to oxidize it, must be HOT and CONCENTRATED. Just putting Cu in some H2SO4 will not produce a reaction. The reaction of Cu + H2SO4 is as follows: Cu(s) + 2H2SO4(l) ---> Cu2SO4(s) + SO2(g) + 2H2O(g) ... note the states of each. In this reaction, the S of H2SO4 has gained electrons to form SO2, so it has been reduced and it has oxidized the Cu which went from Cu(s) to Cu^2+ in CuSO4. HCl is not capable of gaining electrons from Cu so it cannot oxidize the Cu.
H, Mg, Zn, Cu
When H2SO4 dissociates in water, it forms two steps of dissociation. First, it breaks into H+ and HSO4-. Then, HSO4- further dissociates into H+ and SO4^2-. This results in the formation of 2 moles of H+ ions and 1 mole of SO4^2- ion for every mole of H2SO4 dissociated.
The word equation for Cu + H2SO4 is copper + sulfuric acid yields copper(II) sulfate + hydrogen gas.
The elements in H2SO4 are hydrogen (H), sulfur (S), and oxygen (O).
The balanced chemical equation for the reaction between copper hydroxide (Cu(OH)2) and sulfuric acid (H2SO4) is: Cu(OH)2 + H2SO4 → CuSO4 + 2H2O
It is exothermic. Take for example H2SO4 H2SO4 -> H+ + HSO4- This is very exothermic