Chemistry, 17.09.2019 07:30 nuggetslices

Calculate the standard emf of a cell that uses the mg/mg2+ and cu/cu2+ half-cell reactions at 25 °c. write the equation for the cell reaction that occurs under standard-state conditions and write the line notation for the cell.

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Answer from: Aethis

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 + Mg
Half cell reactions:
Cu > Cu2 + 2e-
Mg2 + 2e- > Mg
Standard potentials at 25 degrees Celsius
Cu > Cu2 + 2e- (E = +0.35 V)
Mg2 + 2e- > Mg (E = -2.36 V)

Adding the standard potentials, then we will have
+0.35 V + (-2.36 V) = -2.01 V

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.

Answer from: sksksjs

The cell reaction that occurs is as follows:

$\boxed{{\text{Mg + C}}{{\text{u}}^{{\text{2 + }}}} \rightleftarrows {\text{M}}{{\text{g}}^{2 + }}{\text{ + Cu}}}$

The line notation of cell is as follows:

$\left. {{\text{Mg}}\left( {\text{s}} \right)} \right|{\text{M}}{{\text{g}}^{2 + }}\left( {{a_{{\text{M}}{{\text{g}}^{2 + }}}}} \right)\left\| {{\text{C}}{{\text{u}}^{2 + }}\left( {{a_{{\text{C}}{{\text{u}}^{2 + }}}}} \right)\left| {{\text{Cu}}\left( {\text{s}} \right)} \right.} \right.$

The standard emf value of the cell is $\boxed{2.7{\text{ V}}}$ .

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,

${\text{X}} + {\text{Y}} \to {{\text{X}}^ + }+{{\text{Y}}^ - }$

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.

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 ${\text{Mg/M}}{{\text{g}}^{2 + }}$ is $+ 2.363{\text{ V}}$ .

The standard oxidation potential for ${\text{Cu/C}}{{\text{u}}^{2 + }}$ is $- 0.337{\text{ V}}$ .

Since ${\text{Mg}}$ has higher oxidation potential thus the oxidation of ${\text{Mg}}$ takes place at anode and reduction of ${\text{C}}{{\text{u}}^{2 + }}$ takes place at cathode.

The oxidation half-reaction of ${\text{Mg/M}}{{\text{g}}^{2 + }}$ can be written as:

${\text{Mg}} \to {\text{M}}{{\text{g}}^{2 + }} + 2{e^ - }$ ......(1)

The reduction half-reaction ${\text{Cu/C}}{{\text{u}}^{2 + }}$ can be written as:

${\text{C}}{{\text{u}}^{2 + }} + 2{e^ - } \to {\text{Cu}}$ ......(2)

Add reaction (1) and (2) and eliminate common terms to determine the net reaction for the given cell.

${\text{Mg + C}}{{\text{u}}^{{\text{2 + }}}} \rightleftarrows {\text{M}}{{\text{g}}^{2 + }}{\text{ + Cu}}$ ......(3)

The expression of the cell is written as follows:

The expression to calculate the standard emf of the cell is as follows:

$E_{{\text{cell}}}^0 = E_{{\text{anode}}}^0 - E_{{\text{cathode}}}^0$ ......(5)

Substitute $+ 2.363{\text{ V}}$ for $E_{{\text{anode}}}^0$ and $- 0.337{\text{ V}}$ for $E_{{\text{cathode}}}^0$ in equation (5).

$\begin{aligned}E_{{\text{cell}}}^0&=\left({ + 2.363{\text{ V}}}\right)-\left( { - 0.337{\text{ V}}} \right)\\&= 2.7{\text{ V}}\\\end{aligned}$

Learn more:

1. Which occurs during the redox reaction?

2. Oxidation and reduction reaction:

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.

Answer from: Quest

where are the choice options?

Answer from: Quest

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and conduct electricity easily while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.

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Calculate the standard emf of a cell that uses the mg/mg2+ and cu/cu2+ half-cell reactions at 25 °c....