In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. The two half-reactions that occur in the cell are Cu2 (aq) 2e−→Cu(s) and Co(s)→Co2 (aq) 2e− The net reaction is Cu2 (aq) Co(s)→Cu(s) Co2 (aq) Use the given standard reduction potentials in your calculation as appropriate.

Respuesta :

Answer: The [tex]E^o_{cell}\text{ and }K_{eq}[/tex] of the reaction is 0.62 V and [tex]9.55\times 10^{20}[/tex]

Explanation:

We are given:

Oxidation half reaction:  [tex]Co(s)\rightarrow Co^{2+}(aq.)+2e^-[/tex]   [tex]E^o_{Co^{2+}/Co}=-0.28V[/tex]

Reduction half reaction:  [tex]Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)[/tex]   [tex]E^o_{Cu^{2+}/Cu}=0.34V[/tex]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

Putting values in above equation, we get:

[tex]E^o_{cell}=0.34-(-0.28)=0.62V[/tex]

To calculate the [tex]K_{eq}[/tex] of the reaction, we use Nernst equation, which is:

[tex]E^o_{cell}=\frac{0.0591}{n}\log K_{eq}[/tex]

where,

[tex]E^o_{cell}[/tex] = standard electrode potential = 0.62 V

n = number of electrons transferred = 2

Putting values in above equation, we get:

[tex]0.62=\frac{0.0591}{2}\log K_{eq}\\\\K_{eq}=9.55\times 10^{20}[/tex]

Hence, the [tex]E^o_{cell}\text{ and }K_{eq}[/tex] of the reaction is 0.62 V and [tex]9.55\times 10^{20}[/tex]

The branch of chemistry which deals with the making of electricity with the help of chemical reactions is called electrochemistry.

The correct answer is 0.62 [tex]9.55*10^{20[/tex].

REDOX REACTION

  • Redox is a type of chemical reaction in which the oxidation states of atoms are changed.
  • Redox reactions are characterized by the actual or formal transfer of electrons between chemical species

The oxidation potential is as follows:- [tex]E^o_{Co^{2+}/Co = -0.28\\ [/tex]

The reduction potential is as follows:- [tex]E^o_{Cu^{2+}/Cu=0.34[/tex]

The potential of the cell is as follows:-

[tex]E^o_cell =0.34-(-0.28) =0.62V[/tex].

Let's use the Nerst equation to find the keq?

[tex]E^o_{cell} = \frac{0.0591}{n}logK_{eq}[/tex],

Placed all the values to the equation.

[tex]0.62= \frac{0.0591}{2}logK_{eq}\\ \\ K_{eq} =9.55*10^{20}[/tex]

Hence, the correct answer is mentioned above.

For more information about the question, refer to the link:-

https://brainly.com/question/9552459