acetic acid + copper
CH3COOH + Cu -> CuCH3COOH?
hyrdrogen peroxide (catalyzed by manganese dioxide)
H2)2 -(MnO2)> H2 + O2?
thanks in advance
1. no. Cu is below H in the activity series and there will be no reaction between Cu and acetic acid.
2. looks ok to me.
for number one, do you mean it is impossible for copper to be after the acid?
I mean that Cu and CH3COOH do not react. For all practical purposes, they sit there and look at each other. If you put copper shavings in acetic acid and leave it for several days, you will get a small amount of green copper acetate but the amount is small and it takes some time to get even a little. Compare this with a metal, such as zinc and the reaction with an acid; there are volumes and volumes of H2 gas formed.
The chemical reaction you mentioned involves acetic acid (CH3COOH) and copper (Cu), as well as hydrogen peroxide (H2O2) catalyzed by manganese dioxide (MnO2). Let's break down each reaction and see if the given equations are correct:
1. Acetic acid + Copper:
The reaction between acetic acid and copper involves a redox reaction. Acetic acid is a weak acid and copper is a metal. When copper reacts with acetic acid, it displaces hydrogen from the acid.
The correct balanced equation for this reaction is:
2 CH3COOH + 2 Cu -> Cu(CH3COO)2 + H2
The balanced equation indicates that two molecules of acetic acid react with two atoms of copper, producing one molecule of copper(II) acetate (Cu(CH3COO)2) and one molecule of hydrogen gas (H2).
2. Hydrogen peroxide (Catalyzed by manganese dioxide):
When hydrogen peroxide is catalyzed by manganese dioxide, it decomposes into water and oxygen gas. Manganese dioxide acts as a catalyst, which speeds up the reaction without being consumed in the process.
The correct balanced equation for this reaction is:
2 H2O2 (catalyzed by MnO2) -> 2 H2O + O2
The balanced equation shows that two molecules of hydrogen peroxide decompose into two molecules of water and one molecule of oxygen gas, with the assistance of manganese dioxide as a catalyst.
I hope this helps! If you have any further questions, please feel free to ask.