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monitta
2 years ago
15

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

Chemistry
2 answers:
olga nikolaevna [1]2 years ago
7 0

Reduction refers to a chemical reaction in which electrons are gained by one of the atoms involved in the reaction and at the same time lowering of an oxidation state of that atom takes place.  

In the given case, the reduction occurs at the cathode. In aqueous, V (vanadium) is present in +5 oxidation state that on reduction modifies to vanadium (aq) with +1 oxidation state.  

The half reaction is:  

V⁵⁺ (aq) + 4e⁻ → V¹⁺ (aq)


stealth61 [152]2 years ago
6 0

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<em> gain of electrons</em> or the decrease in the oxidation state of the atom is called reduction while that of <em>loss of electrons </em>or the increase in the<em> oxidation</em> 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? brainly.com/question/1616320

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

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.

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2 years ago
4.82 g of an unknown metal is heated to 115.0∘C and then placed in 35 mL of water at 28.7∘C, which then heats up to 34.5∘C. What
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<u>Answer:</u> The specific heat of metal is 2.34 J/g°C

<u>Explanation:</u>

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When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of metal = 4.82 g

m_2 = mass of water = 35 g

T_{final} = final temperature = 34.5°C

T_1 = initial temperature of metal = 115°C

T_2 = initial temperature of water = 28.7°C

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c_2 = specific heat of water = 4.186 J/g°C

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4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)]

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