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Dahasolnce [82]
3 years ago
13

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

Chemistry
2 answers:
andrew-mc [135]3 years ago
8 0

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

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

<span> 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.</span>

Xelga [282]3 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 <em>gain of electrons </em>or the decrease in the oxidation state of the atom is called <em>reduction </em>while that of <em>loss of electrons </em>or the increase in the oxidation number is known as <em>oxidation</em>. 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 0to +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^-}

Learn more:

1. Which occurs during the redox reaction? brainly.com/question/1616320

2. Oxidation and reduction reaction: brainly.com/question/2973661

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Redox reactions

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

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Answer is: <span>yield of a reaction is 56,4%.
</span>Chemical reaction: PCl₃ + 3H₂O → 3HCl + H₃PO₃.
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abruzzese [7]

Silver chloride produced : = 46.149 g

Limiting reagent : CuCl2

Excess remains := 3.74 g

<h3>Further explanation</h3>

Reaction

silver nitrate + copper(II) chloride ⇒ silver chloride + copper(II) nitrate

Required

silver chloride produced

limiting reagent

excess remains

Solution

Balanced equation

2AgNO3 (aq) + CuCl2 (s) → 2AgCl(s) + Cu(NO3)2(aq)

mol AgNO3 :

= 58.5 : 169,87 g/mol

= 0.344

mol CuCl2 :

=21.7 : 134,45 g/mol

= 0.161

mol ratio : coefficient of AgNO3 : CuCl2 :

= 0.344/2 : 0.161/1

= 0.172 : 0.161

CuCl2  as a limiting reagent

mol AgCl :

= 2/1 x 0.161

= 0.322

Mass AgCl :

= 0.322 x 143,32 g/mol

= 46.149 g

mol remains(unreacted) for AgNO3 :

= 0.344-(2/1 x 0.161)

= 0.022

mass AgNO3 remains :

= 0.022 x 169,87 g/mol

= 3.74 g

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