Example 1: Acidified K₂Cr₂O₇ with H₂O₂

In this reaction, an acidified solution of Potassium DiChromate was reacted with hydrogen peroxide, resulting in the formation of a green-coloured solution. The following half reactions show what is happening in the pictures to the right.

Reduction Half Reaction: Cr₂O₇^2- + 14H⁺ + 6e^- --> 2Cr³⁺ + 7H₂O Eºr = +1.23 V

Oxidation Half Reaction: 3 [H₂O₂ --> 3O₂ + 2H⁺ + 2e^-] Eºr = - 0.70 V

Balanced Net Equation: Cr₂O₇^2- + 8H⁺ + 3H₂O₂ --> 2Cr³⁺ + 7H₂O + 3O₂ Eº = +0.53V

The reaction is spontaneous; not only is the theoretical voltage positive, but as you can see from the pictures to the right, there is evidence of a reaction without any external source of energy.

Descriptions of reactants and products:

K₂Cr₂O₇ - liqud, yellow, clear, odourless

H₂O₂ - liquid, colourless, clear, odourless

Product - liquid, olive green, clear, odourless

Identification of product: From the net equation, the products are Cr³⁺, H₂O and O₂. Cr³⁺ ion gives an olive green colour in aqueous solution. This indicates the presence of Cr³⁺ ions.

A bit about Chromium: Chromium (III) ion is required in trace amounts for sugar metabolism in humans. Chromium deficiency during total parenteral nutrition has been shown to cause insulin resistance, resulting in diabetes-like symptoms. This can be reversed by taking chromium supplements. Chromium is widely distributed in the food supply, but most foods provide small amounts. Good sources of chromium include meat, whole-grain products, fruits, vegetables, and spices. The popular dietary supplement chromium picolinate is promoted as an aid to body development for athletes. However recently a number of studies have failed to demonstrate an effect of chromium picolinate on either muscle growth or fat loss. In addition, there is concern that chromium picolinate could cause DNA damage and mutation than other forms of chromium (III) ion. Although these results are still a subject of debate, it is better to get chromium from natural food rather than supplements.

Example 2: Cu with AgNO₃

In this reaction, a copper wire was cleaned with sandpaper to remove any oxides on its surface. The wire was then placed into a solution of silver nitrate. Metalic crystals formed on the copper wire after several minutes. The following analysis explains at an atomic level the reason behind this beautiful transformation.

Reduction Half Reaction: 2 [Ag⁺ + e^- --> Ag] Eºr = + 0.80V

Oxidation Half Reaction: Cu --> Cu²⁺ + 2e^- Eºr = - 0.34V

Balanced Net Equation : 2Ag⁺ + Cu --> 2Ag + Cu²⁺ Eº = +0.46V

The reaction is spontaneous; the voltage is positive, and the silver crystals that appear in the pictures to the left do so without any energy being added.

Descriptions of reactants and products:

Cu - solid, copper-coloured, lustrous, malleable

AgNO₃ - liquid, colourless, clear, odourless

Product - Silvery-coloured lustrous flakes can be seen wrapping around the copper wire. The flakes can be shaken off easily. The solution turned to a clear blue colour.

Identification of product: From the net equation, the final products are Ag and Cu²⁺. The flakes are silver flakes, indicating the presence of silver. Cu²⁺ ion gives a blue colour in aqueous solution. The blue solution indicates the presence of Cu²⁺ ions.

Significant comments: While this reaction can be used to isolate silver from silver nitrate, it is not a common process used to extract silver in real practice. Silver nitrate is the most important compound of silver. It is a very toxic compound if in taken. It is used in the preparation of silver salts for photography because of its light-sensitive characteristic. It is also used in preparation in silver plating, chemical analysis, in inks and hair dyes, and to silver mirrors. In medicine, it is used to treat eye infections and gonorrhea in newborns for its antiseptic properties. In histology, silver nitrate is used for silver staining for demonstrating proteins and nucleic acids. For this reason it is also used to demonstrate proteins in PAGE gels. It is also used as a stain in scanning electron microscopy.

Example 3: Acidified KMnO₄ with solid Iron (II) Sulphate

In this reaction, an acidified solution of Potassium Permanganate was reacted with solid Iron (II) Sulphate crystals, resulting in a clear solution. To find out why the crystals disappear and the purple solution changes colour, check out the chemistry below!

Reduction Half Reaction: MnO₄^- + 8H⁺ + 5e^---> Mn²⁺ + 4H₂O Eºr = + 1.51V

Oxidation Half Reaction: 5[Fe²⁺ -->Fe³⁺ + e^-] Eºr = - 0.77V

Balanced Net Equation : MnO₄^- + 8H⁺ + 5Fe²⁺ --> 5Fe³⁺ + Mn²⁺ + 4H₂O

Eº = +0.74V

The reaction is spontaneous; the difference in reduction potentials is positive, and the reaction proceded instantaneously when the reactants were combined.

Descriptions of reactants and products:

FeSO₄- solid, green/teal, dull, crystal-like and brittle

KMnO₄ - liquid, purple, odorless

Product - Clear solution, no sediment or solids.

Identification of product: From the net equation, the final products are Fe³⁺ and Mn²⁺ ions. Both ions are colourless, so the solution supports this predition. In addition, the purple colour disappeared, suggesting that the MnO₄,^- ions are no longer present.

MnO₄^-'s applications : The reaction featured above could be desribed alternatively as the oxidataion of Iron by Potassium Permanganate. In fact, the oxidation of substances by Potassium Permanganate is a very common reaction, used often in the treatment of municiple water systems. The reason for this is based in the position of the MnO₄^- ion on our redox table: it is one of the strongest oxidizing, agents, which means that it can oxidize almost any other substance. This allows treatment centers to oxidize contaminants such as metals (for example, iron!), or organic substances. The Mn²⁺ ion spontaneously reacts with dissolved O_2(g) in the water to form MnO_2(s)

Example 4: Acidified KMnO₄ with H₂O₂

In this reaction, an acidified solution of Potassium Permanganate was reacted with liquid hydrogen peroxide, resulting in a clear solution; during the reaction, bubbles were observed, as can be seen in the second picture. How did it happen? The analysis below explains how the transfer of electrons between the reactants resulted in this clear solution.

Reduction Half Reaction: 2[MnO₄^- + 8H⁺ + 5e^---> Mn²⁺ + 4H₂O] Eºr = + 1.51V

Oxidation Half Reaction: 5[H₂O₂ --> O₂ + 2H⁺ + 2e^-] Eºr = - 0.70V

Balanced Net Equation : 2MnO₄^- + 6H⁺ + 5H₂O₂ --> 5O₂ + 2 Mn²⁺ + 8H₂O

Eº = +0.81V

The reaction is spontaneous; the difference in reduction potentials is positive, and the reaction proceded quickly on its own with the introduction of the hydrogen peroxide reactant.

Descriptions of reactants and products:

H₂O₂- clear, colourless liquid, odorless

KMnO₄ - liquid, purple, odorless

Product - Clear solution, no sediment or solids; during reaction, a large quantity of bubbles were observed.

Identification of product: From the net equation, the final products are water and Mn²⁺ ions. Both are colourless, so the solution supports this predition. In addition, bubbles were observed, suggesting that the equation is correct in its assumption that oxygen gas was released.

Environmental advantages of H₂O₂: Hydrogen peroxide's #1 use is in the bleaching of paper; however, it still lags far behind other bleachers, such as Chlorine. In the last couple decades, more and more people have began to demand that the paper they use is 'environmentally friendly'. Since chlorine bleaching leaves highly toxic waste products, many of which end up in rivers where they cause harm to the environment, many people and 'E-friendly corporations' will no longer purchase chlorine treated paper. An alternative that many pulp and paper mills use to bleach the paper is hydrogen peroxide. Environmentally, hydrogen peroxide has significant advantages, because it decomposes to hydrogen and oxygen gas quite quickly in water, thus making it safe to release into rivers; this is attractive to the manufacturer, because they don't have to put in expensive waste treatment plants to please their consumers.