Question:
\[KMn{{O}_{4}}\] spot can be bleached by which of the following:
a.) \[{{H}_{2}}{{O}_{2}}/{{H}^{+}}\]
b.) \[S{{O}_{2}}/{{H}^{+}}\]
c.) \[{{C}_{2}}{{O}_{4}}^{2-}/{{H}^{+}}\]
d.) All of these
Answer
655.5k+ views
Hint: \[KMn{{O}_{4}}\]is an oxidising agent and it oxidises the compound it reacts with thereby reducing itself. \[KMn{{O}_{4}}\] is violet in colour in its oxidised form. To bleach the spot we have to treat it with a compound which can undergo oxidation in its redox reaction with \[KMn{{O}_{4}}\]. The compound among the options which can be oxidised can be used to bleach the spot of \[KMn{{O}_{4}}\].
Complete step by step answer:
Oxidation is the process of loss of electrons during a reaction by a molecule, atom or ion. It occurs when the oxidation state of a molecule, atom or ion is increased. The process opposite to this is called reduction, which occurs when there takes place a gain of electrons or the oxidation state of a species decreases.
\[KMn{{O}_{4}}\] reacts with all of these reagents and oxidise them. In all of these reactions the violet colour of solution disappears due to reduction of permanganate ion in the solution. The manganese element has an oxidation state of +7 in the permanganate ion, when treat with above mentioned reagents it reduces to \[M{{n}^{2+}}\] ions. As we know, \[{{H}_{2}}{{O}_{2}}/{{H}^{+}}\] can be oxidised to dioxygen gas, also \[S{{O}_{2}}/{{H}^{+}}\] can be oxidised to sulphur trioxide or sulphuric acid, and finally \[{{C}_{2}}{{O}_{4}}^{2-}/{{H}^{+}}\] can be easily oxidised to carbon dioxide.
So, the correct answer is “Option D”.
Note: A redox reaction can be defined as a chemical reaction in which electrons are transferred between the two reactants participating in it. This transfer of electrons can be identified by observing the changes taking place in the oxidation states of the reacting species. An oxidising agent is a substance that has the ability to oxidize other substances or to accept their electrons.
Reducing agent is the name given to an element or compound that loses an electron to an electron recipient in a redox chemical reaction. A reducing agent is thus oxidized when it loses electrons in a reaction. Reducing agents reduce the oxidizing agents.
Complete step by step answer:
Oxidation is the process of loss of electrons during a reaction by a molecule, atom or ion. It occurs when the oxidation state of a molecule, atom or ion is increased. The process opposite to this is called reduction, which occurs when there takes place a gain of electrons or the oxidation state of a species decreases.
\[KMn{{O}_{4}}\] reacts with all of these reagents and oxidise them. In all of these reactions the violet colour of solution disappears due to reduction of permanganate ion in the solution. The manganese element has an oxidation state of +7 in the permanganate ion, when treat with above mentioned reagents it reduces to \[M{{n}^{2+}}\] ions. As we know, \[{{H}_{2}}{{O}_{2}}/{{H}^{+}}\] can be oxidised to dioxygen gas, also \[S{{O}_{2}}/{{H}^{+}}\] can be oxidised to sulphur trioxide or sulphuric acid, and finally \[{{C}_{2}}{{O}_{4}}^{2-}/{{H}^{+}}\] can be easily oxidised to carbon dioxide.
So, the correct answer is “Option D”.
Note: A redox reaction can be defined as a chemical reaction in which electrons are transferred between the two reactants participating in it. This transfer of electrons can be identified by observing the changes taking place in the oxidation states of the reacting species. An oxidising agent is a substance that has the ability to oxidize other substances or to accept their electrons.
Reducing agent is the name given to an element or compound that loses an electron to an electron recipient in a redox chemical reaction. A reducing agent is thus oxidized when it loses electrons in a reaction. Reducing agents reduce the oxidizing agents.
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