
The charge required for the oxidation of one mole of$M{n_3}{O_4} \to Mn{O_4}^{2 - }$ in alkaline medium is: (assume $100\%$ current efficiency)
A.$\dfrac{{10}}{3}F$
B.6 F
C.10 F
D.4 F
Answer
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Hint: A chemical reaction is a process that results in the chemical change of one set of chemical substances into another set of chemical substances. Chemical reactions are often defined as changes in the locations of electrons in the formation and breaking of chemical bonds between atoms, with no change in the nuclei (no change in the elements present), and may be represented using a chemical equation.
Complete answer:
The loss of electrons or a rise in the oxidation state of an atom, an ion, or specific atoms in a molecule is referred to as oxidation. The actual or formal transfer of electrons between chemical species is defined by redox reactions, which usually include one species (the reducing agent) suffering oxidation (losing electrons) while another species (the oxidising agent) experiences reduction (gains electrons). The chemical species that loses an electron is said to have been oxidised, whereas the chemical species that gains an electron is said to have been reduced. Although the production of oxides from oxygen molecules is typically linked with oxidation processes, other chemical entities can perform the same role.
In alkaline medium,$M{n_3}{O_4} \to Mn{O_4}^{2 - }$
$M{n_3}^{ + \dfrac{6}{3}}{O_4} + O{H^ - } \to M{n^{ + 6}}{O_4}^{2 - }$
Here the change in charge is represented as $(6 - \dfrac{8}{3}) \times 3$
Upon simplifying we get
= 18 - 8
= 10 F
Hence we can conclude that to decrease 1 mole of\[M{n_3}{O_4}\], a 10 F charge is required.
Hence option C is the correct answer.
Note:
Similar to acid–base reactions, oxidation and reduction occur concurrently and cannot occur independently of one another. Because two half-reactions usually occur together to produce a full reaction, oxidation and reduction are each called a half-reaction. In order to balance the half-reaction in terms of electric charge, the acquired or lost electrons are usually explicitly mentioned when writing half-reactions. When the half-reactions are combined to form the net chemical equation, the electrons cancel out.
Complete answer:
The loss of electrons or a rise in the oxidation state of an atom, an ion, or specific atoms in a molecule is referred to as oxidation. The actual or formal transfer of electrons between chemical species is defined by redox reactions, which usually include one species (the reducing agent) suffering oxidation (losing electrons) while another species (the oxidising agent) experiences reduction (gains electrons). The chemical species that loses an electron is said to have been oxidised, whereas the chemical species that gains an electron is said to have been reduced. Although the production of oxides from oxygen molecules is typically linked with oxidation processes, other chemical entities can perform the same role.
In alkaline medium,$M{n_3}{O_4} \to Mn{O_4}^{2 - }$
$M{n_3}^{ + \dfrac{6}{3}}{O_4} + O{H^ - } \to M{n^{ + 6}}{O_4}^{2 - }$
Here the change in charge is represented as $(6 - \dfrac{8}{3}) \times 3$
Upon simplifying we get
= 18 - 8
= 10 F
Hence we can conclude that to decrease 1 mole of\[M{n_3}{O_4}\], a 10 F charge is required.
Hence option C is the correct answer.
Note:
Similar to acid–base reactions, oxidation and reduction occur concurrently and cannot occur independently of one another. Because two half-reactions usually occur together to produce a full reaction, oxidation and reduction are each called a half-reaction. In order to balance the half-reaction in terms of electric charge, the acquired or lost electrons are usually explicitly mentioned when writing half-reactions. When the half-reactions are combined to form the net chemical equation, the electrons cancel out.
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