Write a balanced half-reaction describing the oxidation of aqueous vanadium(i) cations to aqueous vanadium(v) cations.

Chemistry

2 answers:

andrew-mc [135]2 years ago

8 0

The balanced half-reaction: V⁺(aq) → V⁵⁺(aq) + 4e⁻.

Vanadium(I) cations lose four electrons and became vanadium(V) cations.

Oxidation reaction is increasing of oxidation number of element, because element or ion lost electrons in chemical reaction.
Reduction is lowering oxidation number because element or ions gain electrons.

Xelga [282]2 years ago

4 0

The balanced half-cell reaction for the oxidation of aqueous vanadium (I) cations to aqueous vanadium (V) cations is $\boxed{{{\text{V}}^{1 + }}\left( {aq} \right) \to {{\text{V}}^{5 + }}\left( {aq} \right) + 4{e^ - }}$ .

Further Explanation:

Redox reaction:

It is a type of chemical reaction in which the oxidation states of atoms are changed. In this reaction, both reduction and oxidation are carried out at the same time. Such reactions are characterized by the transfer of electrons between the species involved in the reaction.

The process of gain of electrons or the decrease in the oxidation state of the atom is called reduction while that of loss of electrons or the increase in the oxidation number is known as oxidation. In redox reactions, one species lose electrons and the other species gain electrons. The species that lose electrons and itself gets oxidized is called as a reductant or reducing agent. The species that gains electrons and gets reduced is known as an oxidant or oxidizing agent. The presence of a redox pair or redox couple is a must for the redox reaction.

The general representation of a redox reaction is,

${\text{X}}+{\text{Y}}\to{{\text{X}}^+}+{{\text{Y}}^-}$

The oxidation half-reaction can be written as:

${\text{X}}\to{{\text{X}}^+}+{e^-}$

The reduction half-reaction can be written as:

${\text{Y}}+{e^-}\to{{\text{Y}}^-}$

Here, X is getting oxidized and its oxidation state changes from $0$ to +1 whereas B is getting reduced and its oxidation state changes from 0 to -1. Hence, X acts as the reducing agent whereas Y is an oxidizing agent.

Initially, vanadium is present in +1 oxidation state. Its oxidation state changes from +1 to +5 oxidation state. The oxidation state of V is increased during the reaction so oxidation is taking place. The balanced oxidation half-cell reaction is as follows:

${{\text{V}}^{1+}}\left({aq}\right)\to{{\text{V}}^{5+}}\left({aq}\right)+4{e^-}$