Respuesta :
To determine the standard emf of the given cells, we need to know the half cell reactions and their corresponding standard potentials at a temperature of 25 degrees Celsius. It would be as follows:
The reaction:
Cu + Mg2 ----> Cu2 + MgÂ
Half cell reactions:
Cu --------> Cu2 + 2e-Â
Mg2 + 2e- -----> MgÂ
Standard potentials at 25 degrees Celsius
Cu -----> Cu2 + 2e- Â Â Â (E = +0.35 V)Â
Mg2 + 2e- -----> Mg   (E = -2.36 V)Â
Adding the standard potentials, then we will have
+0.35 V + (-2.36 V) = -2.01 VÂ
Therefore, the standard emf of the redox reaction Cu + Mg2 ----> Cu2 + Mg  would be equal to -2.01 V. The negative sign indicates the equilibrium of the reaction lies to the left of the equation.
The reaction:
Cu + Mg2 ----> Cu2 + MgÂ
Half cell reactions:
Cu --------> Cu2 + 2e-Â
Mg2 + 2e- -----> MgÂ
Standard potentials at 25 degrees Celsius
Cu -----> Cu2 + 2e- Â Â Â (E = +0.35 V)Â
Mg2 + 2e- -----> Mg   (E = -2.36 V)Â
Adding the standard potentials, then we will have
+0.35 V + (-2.36 V) = -2.01 VÂ
Therefore, the standard emf of the redox reaction Cu + Mg2 ----> Cu2 + Mg  would be equal to -2.01 V. The negative sign indicates the equilibrium of the reaction lies to the left of the equation.
The cell reaction that occurs is as follows:
 [tex]\boxed{{\text{Mg + C}}{{\text{u}}^{{\text{2 + }}}} \rightleftarrows {\text{M}}{{\text{g}}^{2 + }}{\text{ + Cu}}}[/tex]
The line notation of cell is as follows:
[tex]\left. {{\text{Mg}}\left( {\text{s}} \right)} \right|{\text{M}}{{\text{g}}^{2 + }}\left( {{a_{{\text{M}}{{\text{g}}^{2 + }}}}} \right)\left\| {{\text{C}}{{\text{u}}^{2 + }}\left( {{a_{{\text{C}}{{\text{u}}^{2 + }}}}} \right)\left| {{\text{Cu}}\left( {\text{s}} \right)} \right.} \right.[/tex]
The standard emf value of the cell is [tex]\boxed{2.7{\text{ V}}}[/tex].
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 general representation of a redox reaction is,
 [tex]{\text{X}} + {\text{Y}} \to {{\text{X}}^ + }+{{\text{Y}}^ - }[/tex]
The oxidation half-reaction can be written as:
[tex]{\text{X}} \to {{\text{X}}^ + } + {e^ - }[/tex]
The reduction half-reaction can be written as:
[tex]{\text{Y}} + {e^ - } \to {{\text{Y}}^ - }[/tex] Â
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.
The element which has higher oxidation potential is oxidized at anode and the element with the less oxidation potential is reduced at cathode in the cell.
The standard oxidation potential for [tex]{\text{Mg/M}}{{\text{g}}^{2 + }}[/tex] is  [tex]+ 2.363{\text{ V}}[/tex].
The standard oxidation potential for [tex]{\text{Cu/C}}{{\text{u}}^{2 + }}[/tex] is [tex]- 0.337{\text{ V}}[/tex].
Since [tex]{\text{Mg}}[/tex] has higher oxidation potential thus the oxidation of [tex]{\text{Mg}}[/tex] takes place at anode and reduction of [tex]{\text{C}}{{\text{u}}^{2 + }}[/tex] takes place at cathode.
The oxidation half-reaction of [tex]{\text{Mg/M}}{{\text{g}}^{2 + }}[/tex] can be written as:
[tex]{\text{Mg}} \to {\text{M}}{{\text{g}}^{2 + }} + 2{e^ - }[/tex] Â Â ......(1)
The reduction half-reaction [tex]{\text{Cu/C}}{{\text{u}}^{2 + }}[/tex] can be written as:
[tex]{\text{C}}{{\text{u}}^{2 + }} + 2{e^ - } \to {\text{Cu}}[/tex] Â Â ......(2)
Add reaction (1) and (2) and eliminate common terms to determine the net reaction for the given cell.
[tex]{\text{Mg + C}}{{\text{u}}^{{\text{2 + }}}} \rightleftarrows {\text{M}}{{\text{g}}^{2 + }}{\text{ + Cu}}[/tex] Â Â Â Â Â ......(3)
The expression of the cell is written as follows:
[tex]\left. {{\text{Mg}}\left( {\text{s}} \right)} \right|{\text{M}}{{\text{g}}^{2 + }}\left( {{a_{{\text{M}}{{\text{g}}^{2 + }}}}} \right)\left\| {{\text{C}}{{\text{u}}^{2 + }}\left( {{a_{{\text{C}}{{\text{u}}^{2 + }}}}} \right)\left| {{\text{Cu}}\left( {\text{s}} \right)} \right.} \right.[/tex] Â Â Â Â Â Â Â Â Â Â Â Â Â Â ......(4)
The expression to calculate the standard emf of the cell is as follows:
[tex]E_{{\text{cell}}}^0 = E_{{\text{anode}}}^0 - E_{{\text{cathode}}}^0[/tex] Â Â Â Â Â Â Â Â Â ......(5)
                             Â
Substitute [tex]+ 2.363{\text{ V}}[/tex] for [tex]E_{{\text{anode}}}^0[/tex] and [tex]- 0.337{\text{ V}}[/tex] for [tex]E_{{\text{cathode}}}^0[/tex] in equation (5).
[tex]\begin{aligned}E_{{\text{cell}}}^0&=\left({ + 2.363{\text{ V}}}\right)-\left( { - 0.337{\text{ V}}} \right)\\&= 2.7{\text{ V}}\\\end{aligned}[/tex] Â Â Â Â
Learn more:
1. Which occurs during the redox reaction? https://brainly.com/question/1616320
2. Oxidation and reduction reaction: https://brainly.com/question/2973661
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Electrochemistry
Keywords: Mg/M2+, Cu/Cu2+, half-cell reaction, standard emf of the cell, oxidation state, reduction, oxidation, redox reaction, transfer of electrons, reducing agents, oxidizing agents.
