Chemistry, 03.07.2019 08:30 lilGi0

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

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Answer from: aek02

Explanation:

Reduction is a chemical reaction in which electrons are gained by one of the atoms taking part in the reaction and lowering of an oxidation state of that atom.

Reduction takes place at the cathode.

In aqueous, vanadium(V) is present in +5 oxidation state which on reduction changes to vanadium(I) with +1 oxidation state.

The half reaction is :

$V^{5+}(aq)+4e^- \rightarrow V^{1+}(aq)$

Answer from: junior1391

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

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 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 +5 oxidation state. Its oxidation state changes from +5 to +1 oxidation state. The oxidation state of V is decreased during the reaction so reduction is taking place. The balanced reduction half-cell reaction is as follows

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

Learn more:

1. Which occurs during redox reaction?

2. Oxidation and reduction reaction:

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Redox reactions

Keywords: V5+, V1+, 4e-, oxidation state, reduction, oxidation, redox reaction, transfer of electrons, reducing agents, oxidizing agents.

Answer from: Quest

Idon't know hope someone you

Answer from: Quest

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o

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