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Svetlanka [38]
3 years ago
14

Onsider a galvanic cell in which al3 is reduced to elemental aluminum, and magnesium metal is oxidized to mg2 . write the balanc

ed half-cell reactions that take place at the cathode and at the anode.
Chemistry
2 answers:
jok3333 [9.3K]3 years ago
8 0
Two Half Cell reactions are as follow,

                                    3e⁻  +    Al⁺³    →  Al            (Reduction Reaction)

                                    Mg   →   Mg⁺²  +  2e⁻           (Oxidation Reaction)

In order to balance the number of electrons, multiply reduction reaction by 2 and oxidation reaction by 3.
So,
                                    6e⁻  +    2 Al⁺³    →  2 Al           

                                    3 Mg   →   3 Mg⁺²  +  6e⁻           (e⁻ cancelled)
                                _______________________

                               2 Al⁺³ + 3 Mg  →  2 Al  +  3 Mg⁺²

Result:
            At Anode
Magnesium metal is Oxidized and At Cathode Aluminium Ions are reduced.
zalisa [80]3 years ago
5 0

The balanced reduction half-cell reaction at cathode is \boxed{{\text{A}}{{\text{l}}^{3+}}+3{e^-}\to{\text{Al}}}  

The balanced oxidation half-cell reaction at anode is \boxed{{\text{Mg}}\to{\text{M}}{{\text{g}}^{2+}}+2{e^-}}  

Further Explanation:

Redox reaction:

Redox is a term that is used collectively for the reduction-oxidation 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 simultaneously. 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 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.

Reduction occurs at cathode whereas oxidation takes place at anode. So {\text{A}}{{\text{l}}^{3+}}  is reduced to form Al at cathode and Mg is oxidized to {\text{M}}{{\text{g}}^{2+}}  at anode.

The balanced reduction half-cell reaction at cathode is,

{\text{A}}{{\text{l}}^{3+}}+3{e^-}\to{\text{Al}}

The balanced oxidation half-cell reaction at anode is,

{\text{Mg}}\to{\text{M}}{{\text{g}}^{2+}}+2{e^-}

Learn more:

1. Which occur during redox reaction? <u>brainly.com/question/1616320</u>

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

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Redox reactions

Keywords: redox reaction, Mg, Mg2+, Al, Al3+, 2e-, 3e-, cathode, anode, half-cell reaction, oxidation, reduction, reductant, oxidant, reducing agent, oxidizing agent, electrons, redox pair, redox couple, oxidation state, oxidized, reduced, simultaneously, increase, decrease.

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Answer:

The answer to your question is:   C₃H₃O  This is my answer.

Explanation:

Data

Sample = 1.3109 g

CxHyOz

CO₂ = 3.2007 g

H₂O = 1.3102 g

Empirical formula = ?

MW CO2 = 44 g

MW H2O = 18 g

For Carbon

                                     44 g -------------------- 12 g

                                     3.2007 g ------------    x

                                      x = (3.2007 x 12) / 44

                                      x = 0.8729 g of Carbon

                                     12 g of C --------------  1 mol

                                     0.8729 g --------------  x

                                     x = (0.8729 x 1) / 12

                                     x = 0.0727 mol of Carbon

For Hydrogen

                                 18 g ---------------------- 1 g

                              1.3102 g -------------------  x

                                   x = (1.3102 x 1) / 18

                                  x = 0.0727 g of Hydrogen                                      

                                  1 g ------------------------ 1 mol

                                  0.0727g ----------------  x

                                  x = (0.0727 x 1)/1

                                  x = 0.0727 mol of Hydrogen

For oxygen

                    g of Oxygen = g of sample - g of Carbon - g of hydrogen

                    g of Oxygen = 1.3109 - 0.8709 - 0.0727

                    g of Oxygen = 0.3673

                                  16 g of Oxygen ------------- 1 mol of O

                                  0.3673 g ---------------------   x

                                   x = (0.3673 x 1)/ 16

                                   x = 0.0230 mol of Oxygen

Divide by the lowest number of moles

Carbon              0.0727 / 0.023  = 3.1  ≈ 3

Hydrogen         0.0727 / 0.023 = 3.1  ≈ 3

Oxygen             0.0230 / 0.023 = 1

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8 0
3 years ago
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