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Consider the reaction corresponding to a voltaic cell and its standard cell potential. Z n ( s ) + C u 2 + ( a q ) ⟶ C u ( s ) + Z n 2 + ( a q ) Zn(s)+CuX2+(aq)⟶Cu(s)+ZnX2+(aq) E o cell = 1.1032 V Ecello=1.1032 V What is the cell potential for a cell with a 2.700 M solution of Z n 2 + ( a q ) ZnX2+(aq) and 0.1448 M solution of C u 2 + ( a q ) CuX2+(aq) at 436.2 K?C

Respuesta :

Answer : The cell potential of the cell is, 1.079 V

Solution :

The given balanced cell reaction will be,

[tex]Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)[/tex]

Here zinc (Zn) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

Using Nernest equation :

[tex]E_{cell}=E^o_{cell}-\frac{RT}{nF}\log \frac{[Zn^{2+}]^2}{[Cu^{2+}]}[/tex]

where,

R = gas constant = 8.314 J/K.mole

F = Faraday constant = 96500 C

T = temperature = 436.2 K

n = number of electrons in oxidation-reduction reaction = 2

[tex]E_{cell}[/tex] = cell potential of the cell = ?

[tex]E^o_{cell}[/tex] = standard electrode potential = 1.1032 V

concentration of [tex]Zn^{2+}[/tex] = 2.700 M

concentration of [tex]Cu^{2+}[/tex] = 0.1448 M

Now put all the given values in the above equation, we get

[tex]E_{cell}=1.1032-\frac{(8.314)\times (436.2)}{2\times 96500}\log \frac{2.700}{0.1448}=1.079V[/tex]

Therefore, the cell potential of the cell is, 1.079 V

The study of chemicals and bonds is called chemistry.

The correct answer is 1.079 V

What is electrochemistry?

  • The study of chemicals which give electricity is called electrochemistry.

The given balanced cell reaction is as follows:-

[tex]Zn + Cu^{2+}------->Zn^{2+}+Cu[/tex]

Here zinc (Zn) undergoes oxidation by loss of electrons, thus acting as the anode. Copper (Cu) undergoes reduction by a gain of electrons and thus acts as a cathode.

According to Nernst equation :

[tex]E_{cell} = E^o_{cell}-\frac{RT}{nF} log\frac{[Zn^{2+}]^2}{Cu^{2+}} [/tex]

where,

  • R = gas constant = 8.314 J/K.mole
  • F = Faraday constant = 96500 C
  • T = temperature = 436.2 K
  • n = number of electrons in oxidation-reduction reaction = 2
  • standard electrode potential = 1.1032 V
  • concentration of  = 2.700 M
  • concentration of  = 0.1448 M

place all the values to the equation,
[tex]E_{cell} = 1.1032-\frac{8.314*436.2}{2*96500}log\frac{2700}{0.1448}\\ =1.079 [/tex]

Therefore, the cell potential of the cell is, 1.079 V.

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

https://brainly.com/question/25606438

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