Change of Iron(II) Ion to Iron(III) Ion Archives

Redox Reactions by Transfer of Electrons at a Distance

December 1, 2020December 1, 2020 by Veerendra

Redox Reactions by Transfer of Electrons at a Distance

In all redox reactions, electrons are transferred from the reducing agent to the oxidising agent.
When the reducing and oxidising agents are mixed together as in the previous reactions, the transfer of electrons occurs quickly and cannot be detected.
However, when the reducing and oxidising agents are separated by an electrolyte in a U-tube as shown in Figure 3.6, the transfer of electrons occurs through the connecting wires and can be detected by a galvanometer.
The reducing agent loses its electrons and hence undergoes oxidation. The electrode at which electrons are released by the reducing agent is called the negative terminal.
The electrons then flow through the connecting wires to the oxidising agent. The electrode at which electrons are accepted by the oxidising agent is called the positive terminal.
As the oxidising agent accepts the electrons, it undergoes reduction.
The electrolyte allows the movement of ions to take place, thus completing the electric circuit. This ensures a continuous flow of electrons in the external circuit.

When polyatomic ions such as manganate(VII) ion and dichromate(VI) ion are involved in redox reactions, the half-equations are more complex. The following tables illustrate how the half-equations are constructed.

Redox Reactions by Transfer of Electrons at a Distance 2

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Oxidation and reduction in the transfer of electrons at a distance experiment

Aim: To investigate oxidation and reduction in the transfer of electrons at a distance.
Materials: 2.0 mol dm^-3 sulphuric acid, 0.5 mol dm^-3 freshly prepared iron(II) sulphate solution, 0.2 mol dm^-3acidified potassium manganate(VII) solution, 0.5 mol dm^-3 potassium iodide solution, 0.2 mol dm^-3acidified potassium dichromate(VI) solution, 0.2 mol dm^-3 potassium thiocyanate solution, 1% starch solution.
Apparatus: U-tube, galvanometer, connecting wires with crocodile clips, carbon electrodes, retort stand and clamp, test tube, dropper, stoppers with one hole.
Procedure:

A U-tube is clamped to a retort stand.
Dilute sulphuric acid is poured into the U-tube until its levels are 6 cm away from the mouths of the U-tube.
Using a dropper, 0.5 mol dm^-3 iron(II) sulphate solution is carefully added to one of the arms of the U-tube until the layer of iron(II) sulphate solution reaches the height of 3 cm.
In a similar manner as in step 3, 0.2 mol dm^-3 acidified potassium manganate(VII) solution is added to the other arm of the U-tube.
A carbon electrode is placed in each arm of the U-tube.
The electrodes are connected to a galvanometer as shown in Figure. Based on the deflection of the galvanometer, the electrodes that act as the positive terminal and negative terminal are determined.
The set-up is left aside for 30 minutes. Any change is observed.
Using a clean dropper, 1 cm³ of iron(II) sulphate solution is drawn out and placed in a test tube. Then, a few drops of 0.2 mol dm^-3 potassium thiocyanate solution are added to the test tube. Any change is observed.
Steps 1 to 7 are repeated using 0.5 mol dm^-3 potassium iodide solution and 0.2 mol dm^-3 acidified potassium dichromate(VI) solution to replace the iron(II) sulphate solution and acidified potassium manganate(VII) solution. Step 8 is repeated to test the potassium iodide solution with 1 % starch solution.

Results:

1. Solutions used: Iron(II) sulphate solution and acidified potassium manganate(VII) solution

Observation	Inference
(a) The electrode in the iron(II) sulphate solution acts as the negative terminal while the electrode in the acidified potassium manganate(VII) solution acts as the positive terminal.	Electrons flow from iron(II) sulphate solution to acidified potassium manganate(VII) solution.
(b) Iron(II) sulphate solution changes from pale green to yellow. It gives blood-red colouration with potassium thiocyanate solution.	At the end of the reaction, iron(III) ions are present. Iron(II) ions have changed to iron(III) ions.
(c) The purple acidified potassium manganate(VII) solution decolourises.	Manganate(VII) ions that give the solution its purple colour are used up in the reaction.

2. Solutions used: Potassium iodide solution and acidified potassium dichromate (VI) solution

Observation	Inference
(a) The electrode in the potassium iodide solution acts as the negative terminal, whereas the electrode in the acidified potassium dichromate(VI) solution acts as the positive terminal.	Electrons flow from potassium iodide solution to acidified potassium dichromate(VI) solution.
(b) The colourless potassium iodide solution turns brown. It gives a dark blue colouration with starch solution.	At the end of the reaction, iodine is present. Iodide ions have changed to iodine.
(c) Potassium dichromate(VI) solution changes colour from orange to green.	Dichromate(VI) ions have changed to chromium(lll) ions.

Discussion:

1. Iron(II) sulphate solution and acidified potassium manganate(VII) solution
(a) Iron(II) ions act as the reducing agent, releasing electrons to become iron(III) ions. Thus, iron(II) sulphate solution changes colour from pale green to yellow.

(b) The electrons accumulate at the carbon electrode in the iron(II) sulphate solution and flow out to the connecting wires. This carbon electrode acts as the negative terminal.
(c) The electrons then flow to the positive terminal, which is the carbon electrode in the acidified potassium manganate(VII) solution.
(d) Manganate(VII) ions act as the oxidising agent, accepting the electrons and therefore, undergoing reduction to become colourless manganese(II) ions.

(e) The overall ionic equation is as follows:

2. Potassium iodide solution and acidified potassium dichromate(VI) solution
Redox Reactions by Transfer of Electrons at a Distance 7
(a) Iodide ions act as the reducing agent, releasing electrons to become iodine molecules. Thus, the colourless potassium iodide solution turns brown.

(b) The electrons accumulate at the carbon electrode in the potassium iodide solution and flow out to the connecting wires. This carbon electrode acts as the negative terminal.
(c) The electrons then flow to the positive terminal, which is the carbon electrode in the acidified potassium dichromate(VI) solution.
(d) Dichromate(VI) ions act as the oxidising agent, accepting the electrons and therefore, undergoing reduction to become chromium(III) ions.

(e) The overall ionic equation is as follows:

3. The continuous flow of electrons from the reducing agent at the negative terminal to the oxidising agent at the positive terminal produces an electric current that causes the indicator of the galvanometer to deflect.
4. Sulphuric acid has two functions:
(a) To separate the reducing agent from the oxidising agent
(b) To complete the circuit by .allowing the transfer of ions to occur
5. Other electrolytes such as potassium nitrate solution and sodium chloride solution can be used in place of sulphuric acid. The electrolyte should not react with either the reducing agent or the oxidising agent used.
6. There are few other pairs of reducing agent and oxidising agent that can be used in this activity. Here are some examples:
Redox Reactions by Transfer of Electrons at a Distance 11

Conclusion:
The transfer of electrons occurs from the reducing agent to the oxidising agent through the connecting wires.

Redox reaction in the displacement of metals from its salt solution

December 1, 2020December 1, 2020 by Veerendra

Redox reaction in the displacement of metals from its salt solution

Generally, metals are good electron donors and therefore are good reducing agents. However, different metals have different strength as reducing agents.
The strength of metals as reducing agents can be compared by using the electrochemical series.
The electrochemical series lists metals according to their electropositivity, that is, according to their ability to lose electrons to form positive ions.
The higher the position of a metal in the electrochemical series, the more electropositive the metal is, the easier it is for the metal to lose its electrons. Thus, the better reducing agent the metal is.
On the other hand, the ability of a metal ion to accept electrons increases down the series. Thus, the strength of a metal ion as an oxidising agent increases down the electrochemical series.
In a displacement of metal, a more electropositive metal will displace a less electropositive metal from its salt solution.
(a) The more electropositive metal acts as the reducing agent. It loses electrons and undergoes oxidation to form positive ions.
(b) The ions of the less electropositive metal act as an oxidising agent by accepting the electrons. While doing so, the ions are reduced to metallic atoms.
(c) In short, there is an electron transfer from the more electropositive metal to the ions of the less electropositive metal.

Oxidation and reduction in the displacement of metals experiment

Aim: To investigate oxidation and reduction in the displacement of metals.
Materials: Zinc strip, copper strip, 0.5 mol dm^-3 copper(II) sulphate solution, 0.1 mol dm^-3 silver nitrate solution, sandpaper.
Apparatus: Test tubes, test tube rack.
Procedure:

2 cm³ of 0.5 mol dm^-3 copper(II) sulphate solution and 2 cm³ of 0.1 mol dm^-3 silver nitrate solution are poured into two separate test tubes.
A strip of zinc and a strip of copper are cleaned with sandpaper. The strips are then dropped into the test tubes as shown in Figure.
Any change in colour and whether any metal is deposited are observed.

Observations:

Test tube	Observations
X	The blue colour of the solution slowly fades until it becomes colourless. The zinc strip dissolves. A brown solid is deposited.
Y	The colourless solution slowly turns blue. The copper strip dissolves. A silvery grey solid is deposited.

Discussion:

In test tube X, zinc displaces copper from its salt solution.
(a) Zinc is more electropositive than copper. Thus, zinc acts as the reducing agent, losing electrons to form zinc ions. By doing so, zinc is oxidised. This explains why the zinc strip dissolves.

(b) The electrons are accepted by copper(II) ions in the solution. Thus, copper(II) ions act as the oxidising agent and are reduced to metallic copper. The brown solid deposited in test tube X is copper metal.

(c) The decreasing amount of copper(II) ions in the solution causes the solution to slowly change colour from blue to colourless.
(d) The redox reaction that occurs can be represented by the following equation.
In test tube Y, copper displaces silver from its salt solution.
(a) Copper is more electropositive than silver. So, copper acts as the reducing agent, losing electrons to form copper(II) ions. In other words, copper is oxidised. This explains why the copper strip dissolves.

(b) The increasing amount of copper(II) ions in the solution causes the solution to slowly change colour from colourless to blue.

(c) The electrons are accepted by silver ions in the solution. Thus, silver ions act as the oxidising agent and are reduced to silvery grey silver.
(d) The redox reaction that occurs can be represented by the following equation.

Conclusion:
A more electropositive metal can displace a less electropositive metal from its salt solution whereby the more electropositive metal acts as the reducing agent and the ions of the less electropositive metal act as the oxidising agent.

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Changing of iron(II) ions to iron(III) ions and vice versa

December 1, 2020December 1, 2020 by Veerendra

Changing of iron(II) ions to iron(III) ions and vice versa

Iron exhibits two oxidation numbers
(a) +2 as iron(II) ion, Fe²⁺(b) +3 as iron(III) ion, Fe³⁺
An aqueous solution containing iron(II) ions, Fe²⁺ is pale green in colour, whereas that containing iron(III) ions, Fe³⁺is yellow/yellowish-brown/ brown in colour.
Changing iron(II) ions to iron(III) ions is an oxidation and therefore requires an oxidising agent.
On the other hand, changing iron(III) ions to iron(II) ions is a reduction and therefore requires a reducing agent.

Table: Detecting the presence of iron(II) ions and iron(III) ions

Reagent	With iron ions	Observation
Sodium hydroxide solution or ammonia solution	Fe²⁺	Green precipitate, insoluble in excess alkali
Sodium hydroxide solution or ammonia solution	Fe³⁺	Brown precipitate, insoluble in excess alkali
Potassium hexacyanoferrate(ll) solution	Fe²⁺	Light blue precipitation
Potassium hexacyanoferrate(ll) solution	Fe³⁺	Dark blue precipitation
Potassium hexacyanoferrate(lll) solution	Fe²⁺	Dark blue precipitation
Potassium hexacyanoferrate(lll) solution	Fe³⁺	Greenish-brown solution
Potassium/ammonium thiocyanate solution	Fe²⁺	Pale red colouration
Potassium/ammonium thiocyanate solution	Fe³⁺	Blood-red colouration

The following are other oxidising agents that can replace bromine water in changing iron(II) ions to iron(III) ions.
Changing of iron(II) ions to iron(III) ions and vice versa 3

Other reducing agents that can replace zinc powder in changing iron(III) ions to iron(II) ions are as follows.
Changing of iron(II) ions to iron(III) ions and vice versa 4

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Changing of iron(II) ions to iron(III) ions and vice versa experiment

Aim: To investigate oxidation and reduction in the change of iron(II) ions to iron(III) ions and vice versa.
Materials: 0.5 mol dm^-3 freshly prepared iron(II) sulphate solution, 0.5 mol dm^-3 iron(III) sulphate solution, bromine water, zinc powder, 2.0 mol dm^-3 sodium hydroxide solution, filter paper.
Apparatus: Dropper, spatula, test tubes, test tube holder, Bunsen burner, filter funnel, test tube rack.
Procedure:
A. Changing of iron(II) ions to iron(III) ions

2 cm³ of 0.5 mol dm^-3 iron(II) sulphate solution is poured into a test tube.
Using a dropper, bromine water is added to the solution drop by drop.
The test tube is warmed gently.
2.0 mol dm^-3 sodium hydroxide solution is added slowly to the mixture until in excess.

B. Changing of iron(III) ions to iron(II) ions

2 cm³ of 0.5 mol dm^-3 iron(III) sulphate solution is poured into a test tube.
Half a spatula of zinc powder is added to the solution.
The mixture is filtered.
2.0 mol dm^-3 sodium hydroxide solution is added slowly to the filtrate until in excess.

Observations:

Activity	Reagent	Observations
A	Bromine water	Bromine water decolourises. The solution changes colour from pale green to yellow.
A	Sodium hydroxide solution	Brown precipitate is formed. It is insoluble in excess alkali.
B	Zinc powder	Some of the zinc powder dissolves. The solution changes colour from brown to pale green.
B	Sodium hydroxide solution	Green precipitate is formed. It is insoluble in excess alkali.

Discussion:
A. Changing of iron(II) ions to iron(III) ions

Bromine water oxidises iron(II) ions, Fe²⁺ to iron(III) ions, Fe³⁺. The presence of Fe³⁺ ions is confirmed by the formation of brown precipitate with sodium hydroxide solution.
Fe²⁺ ions lose their electrons and are oxidised to Fe³⁺ ions.
Bromine molecules, which give the bromine water its brown colour, gain the electrons and are reduced to colourless bromide ions, Br^–. This explains why the bromine water is decolourised.
In this reaction, bromine water acts as the oxidising agent, where as Fe²⁺ ions act as the reducing agent.

B. Changing of iron(III) ions to iron(II) ions

Zinc powder reduces iron(III) ions, Fe³⁺ to iron(II) ions, Fe²⁺. The presence of Fe²⁺ ions is confirmed by the formation of green precipitate with sodium hydroxide solution.
Zinc atoms lose their electrons and are oxidised to zinc ions, Zn²⁺. This explains why zinc powder dissolves in iron(III) sulphate solution.
Fe³⁺ ions accept these electrons and are reduced to Fe²⁺ ions.
In this reaction, Fe³⁺ ions act as the oxidising agent, whereas zinc acts as the reducing agent.

Conclusion:

Bromine water acts as an oxidising agent, changing iron(II) ions to iron(III) ions.
Zinc acts as a reducing agent, changing iron(III) ions to iron(II) ions.