Classical idea of redox reactions - class-XII

Iron (Fe) and Copper (Cu) are transition metals that frequently participate in redox reactions due to their ability to exist in multiple oxidation states. They are essential cofactors in many biological redox enzymes.

A redox reaction involves the change in oxidation states of atoms. In option B, elemental hydrogen and chlorine react to form hydrogen chloride, where hydrogen is oxidized from 0 to +1 and chlorine is reduced from 0 to -1. The other options represent thermal decomposition or double displacement reactions without changes in oxidation numbers.

A redox reaction is characterized by a change in the oxidation states of the reacting species. Option D is a combination reaction between sulfur trioxide and water to form sulfuric acid, where sulfur remains in the +6 oxidation state, oxygen remains in the -2 state, and hydrogen remains in the +1 state throughout. The other three options involve changes in oxidation states, making them redox processes.

$NO \longrightarrow N _2O$      $E^o= 1.59V$

${ 2I }^{ - }\rightarrow { I } _{ 2 }+{ 2e }^{ - }(oxidation\quad half-reaction)$

${ E } _{ Oxidation }^{ 0 }=-0.536V$

${ Fe }^{ 3+ }+e^{ - }\rightarrow { Fe }^{ 2+ }(reduction\quad half-reaction)$

$E^{ 0 } _{ Reduction }=-0.77V$
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${ 2Fe }^{ 3+ }+{ 2I }^{ - }\rightarrow { 2Fe }^{ 2+ }+{ I } _{ 2 }$

$E^{ 0 }={ E } _{ oxidation }^{ 0 }+E _{ Reduction }^{ 0 },+ve$

Hence the reaction will take place.

${ 2I }^{ - }\rightarrow { I } _{ 2 }+{ 2e }^{ - }(oxidation\quad half-reaction)$

${ E }^{ 0 } _{ Oxidation }=-0.536V$

${ Fe }^{ 3+ }+{ e }^{ - }\rightarrow { Fe }^{ 2+ }(reduction\quad half-reaction)$

$E _{ Reduction }^{ 0 }=-0.77V$
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$2Fe^{ 3+ }+{ 2I }^{ - }\rightarrow { 2Fe }^{ 2+ }+{ I } _{ 2 };$

${ E }^{ 0 }={ E } _{ oxidation }^{ 0 }+{ E } _{ reduction }^{ 0 };+ve$

Oxidation and reduction are two types of chemical reactions that often work together. Oxidation and reduction reactions involve an exchange of electrons between reactants.

In option B, CaO + SiO2 -> CaSiO3, the oxidation states of Ca (+2), O (-2), and Si (+4) remain unchanged. This is an acid-base reaction, not a redox reaction.

Redox reactions involve the transfer of electrons, which is associated with energy changes. These changes can be either exothermic (releasing energy) or endothermic (absorbing energy).

In HN3, the total oxidation state must be 0. Since H is +1, the three nitrogens must sum to -1, resulting in an average oxidation state of -1/3 for each nitrogen atom.

Sulphur dioxide bleaches coloring matter by reduction. Bleaching action of sulphur dioxide is not permanent.

$Mg+CuSO _4 \longrightarrow MgSO _4 + Cu$

In the electrochemical series, metals are arrange in order of their tendency to lose electrons. 

A good oxidizing agent is one which can readily oxidize other chemical species and reduce itself. Therefore, the compound which is good oxidizing agent must have initial oxidation as high, so that it can reduce its oxidation state and get reduced and oxidize others.

$Ans:$ $Silver$ $Nitrate.$

$\begin{array}{l} { H _{ 2 } }{ C _{ 2 } }{ O _{ 4 } }\cdot 2{ H _{ 2 } }O\left( s \right) \to oxadic, acid, dihydrate,  \ Since,, it, is, very, stable, and, also, available, in, pressure \ from, in, the, laboratory, i.e., it's, always, used, as, primary, s\tan  dard, in, the, lab. \ Hence,, the, option, C, is, the, correct, answer. \end{array}$

Balancing the redox reaction Zn + NO3- -> Zn2+ + NH4+ in basic medium involves balancing electrons and charge. The balanced equation is 4Zn + NO3- + 7H2O -> 4Zn2+ + NH4+ + 10OH-. Thus, the coefficients are 4, 1, and 10.

In acidic medium, 2 moles of KMnO4 react with 5 moles of H2O2. Calculating moles of KMnO4 (30.2g / 158g/mol = 0.191 mol), the required moles of H2O2 is (5/2) * 0.191 = 0.4775 mol. Mass = 0.4775 * 34g/mol = 16.2g, which rounds to 16g.

In Redox reaction oxidation state changes