The action of the catalyst on hydrogen peroxide

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Science Investigation.

The action of the catalyst on hydrogen peroxide

Aim: In this investigation I am planning to find out if the volume of gas (O2) is directionally proportional to the varying lengths of potato when the potato is fully submerged in hydrogen peroxide.

Planning

Chemical equation

2H2O2 =2H201+O2

Plan: I am planning to find out how adding H2O2 (hydrogen peroxide) to a piece of potato, reacts with the catalyst, to form oxygen. I could use a gas syringe method or a water trough method.

Water trough method.

  1. Potato
  2. Hydrogen peroxide
  3. Test tube
  4. Delivery tube
  5. Water
  6. Measuring cylinder
  7. Water bath.

In this diagram the potato is in the test tube and so is the hydrogen peroxide. The delivery tube is attached to the test tube and leads to the water bath. The delivery tube then goes under water and faces verticle directly underneath the measuring cylinder. The oxygen then paces through the water into the measuring cylinder. Which in turn makes an air lock . You then measure the oxygen in the air lock in the units cm3. This is the amount of oxygen which has been produced from the reaction.

Gas syringe method.

The gas syringe diagram is in the diagram of apparatus used section.

The two different methods both end up with the same results, but we chose the ‘gas syringe method because we thought it would show a greater accuracy.

Safety

To make this investigation safe I will keep to these rules:

  • Always wear safety goggles
  • Try to all ways stand up when taking part in the experiment.
  • Make sure that all bags, coats folders etc are at the front of the room, or placed fully under the lab tables.
  • If there is a spillage, make sure that you inform a teacher to clear it up.  

Diagram of apparatus used.

                                         


  1. Potato
  2. Hydrogen peroxide
  3. Test tube
  4. Delivery tube
  5. Clamp stand
  6. Gas syringe

Variables

There are several different variables I could change. I could vary the lengths of potato,. This would change the surface area in contact with hydrogen peroxide and the potato.  I could vary the temperature. An increase in temperature would increase the rate of reaction because the heat energy would increase the rate at which the particles moved.

I could of varied the amount of hydrogen peroxide I used. If I used a less amount of pure hydrogen peroxide to the length of potato ie, 4 cm3 of hydrogen peroxide and then a 5 cm piece of potato, the reaction would be slower because, there wouldn’t be the whole surface area of the potato covered by the hydrogen peroxide. Which means not as much oxygen will be produced as if the potato was fully covered by hydrogen peroxide. I could vary the solution of hydrogen peroxide. The weaker the solution of hydrogen peroxide, the less amount of oxygen will be produced, because water does not react with potato to form oxygen. Meaning that the weaker the solution the less oxygen will be produced.

Join now!

I have chosen to vary the lengths of the potato for my variable. I think that this variable will produce the most accurate results, and the most varied results.

 

I will use 5 different lengths of potato. These will be.

  • 1 cm
  • 2 cm
  • 3 cm
  • 4 cm
  • 5 cm

 

Prediction.

I predict that the bigger the cylinders of potato’s are, the bigger amount of oxygen will be produced. I think this will occur because there will be a larger surface area for the hydrogen peroxide to react with, in ...

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