The following reaction takes place in an acid medium:
Write a balanced equation for this reaction.
We need to multiply the right side by two so that the number of Cr atoms will balance. To balance the oxygen atoms, we will need to add water molecules to the right hand side.
Now the oxygen atoms balance but the hydrogens don't. Because the reaction takes place in an acid medium, we can add hydrogen ions to the left side.
The charge on the left of the equation is (-2+14) = +12, but the charge on the right is +6. Therefore, six electrons must be added to the left hand side so that the charges balance. The half reaction is now:
The reduction half reaction after the charges have been balanced is:
We need to multiply the reduction half reaction by three so that the number of electrons on either side are balanced. This gives:
If ammonia solution is added to a solution that contains cobalt(II) ions, a complex ion is formed, called the hexaaminecobalt(II) ion (Co(NH
)
). In a chemical reaction with hydrogen peroxide solution, hexaaminecobalt ions are oxidised by hydrogen peroxide solution to the hexaaminecobalt(III) ion Co(NH
)
. Write a balanced equation for this reaction.
The number of atoms are the same on both sides.
The charge on the left of the equation is +2, but the charge on the right is +3. One elctron must be added to the right hand side to balance the charges in the equation.The half reaction is now:
Although you don't actually know what product is formed when hydrogen peroxide is reduced, the most logical product is OH
. The reduction half reaction is:
After the atoms and charges have been balanced, the final equation for the reduction half reaction is:
We need to multiply the oxidation half reaction by two so that the number of electrons on both sides are balanced. This gives:
Manganate(VII) ions (MnO
) oxidise hydrogen peroxide (H
O
) to oxygen gas. The reaction is done in an acid medium. During the reaction, the manganate(VII) ions are reduced to manganese(II) ions (Mn
). Write a balanced equation for the reaction.
Chlorine gas is prepared in the laboratory by adding concentrated hydrochloric acid to manganese dioxide powder. The mixture is carefully heated.
Write down a balanced equation for the reaction which takes place.
Using standard electrode potentials, show by calculations why this mixture needs to be heated.
Besides chlorine gas which is formed during the reaction, hydrogen chloride gas is given off when the conentrated hydrochloric acid is heated. Explain why the hydrogen chloride gas is removed from the gas mixture when the gas is bubbled through water.
(IEB Paper 2, 2004)
The following equation can be deduced from the table of standard electrode potentials:
(E
= +0.10V)
This equation implies that an acidified solution of aqueous potassium dichromate (orange) should react to form Cr
(green). Yet aqueous laboratory solutions of potassium dichromate remain orange for years. Which ONE of the following best explains this?
Laboratory solutions of aqueous potassium dichromate are not acidified
The E
value for this reaction is only +0.10V
The activation energy is too low
The reaction is non-spontaneous
(IEB Paper 2, 2002)
Sulfur dioxide gas can be prepared in the laboratory by heating a mixture of copper turnings and concentrated sulfuric acid in a suitable flask.
Derive a balanced ionic equation for this reaction using the half-reactions that take place.
Give the E
value for the overall reaction.
Explain why it is necessary to heat the reaction mixture.
The sulfur dioxide gas is now bubbled through an aqueous solution of potassium dichromate. Describe and explain what changes occur during this process.
