In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions: [Zn2+ ] = 0.024 M [H+ ]= 1.3 M partial pressure of H2 = 0.37 atm. Calculate Ecell at 298 K (enter to 3 decimal places). Zn2+ (aq) + 2eLaTeX: -− LaTeX: \longrightarrow⟶ Zn(s) E° = LaTeX: -−0.76 V 2H+ (aq) + 2eLaTeX: -−LaTeX: \longrightarrow⟶ H2(g) E° = 0.00 V

Here Zinc (Zn) undergoes oxidation by loss of electrons, thus act as anode because less value of electrode potential. Hydrogen (H) undergoes reduction by gain of electrons and thus act as cathode because more value of electrode potential.

The half-cell reactions are:

Oxidation reaction (anode) : [tex]Zn(s)\rightarrow Zn^{2+}(aq)+2e^-[/tex]

Reduction reaction (cathode) : [tex]2H^+(aq)+2e^-\rightarrow H_2(g)[/tex]

The balanced cell reaction will be,

[tex]Zn(s)+2H^+(aq)\rightarrow Zn^{2+}(aq)+H_2(g)[/tex]

Given :

[tex]E^o_{[Zn^{2+}/Zn]}=-0.76V[/tex]

[tex]E^o_{[H^+/H_2]}=0.00V[/tex]

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

[tex]E^o=E^o_{[H^+/H_2]}-E^o_{[Zn^{2+}/Zn]}[/tex]

Now put all the values in this expression, we get:

[tex]E^o=0.00V-(-0.76V)=0.76V[/tex]

Now we have to calculate the electrode potential of the cell.

Using Nernst equation :

[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]\times (p_{H_2})}{[H^+]^2}[/tex]

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 298 K

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

[tex]E^o_{cell}[/tex] = electrode potential of the cell = ?

[tex]E_{cell}[/tex] = emf of the cell = 0.76 V

[tex]p_{H_2}[/tex] = 0.37 atm

[tex][Zn^{2+}][/tex] = 0.024 M

[tex][H^{+}][/tex] = 1.3 M

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

[tex]E_{cell}=0.76-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{(0.024}\times (0.37)}{(1.3)^2}[/tex]

[tex]E_{cell}=0.69V[/tex]

Therefore, the electrode potential of the cell is, 0.69 V

abhishekprakash2625 abhishekprakash2625 · Answer 2 · 2025-03-26T09:34:25-04:00

The branch of science which deals with the chemical bond is called chemistry.

The correct answer is 0.69V.

All the data is given in the question and data is as follows:-

Zinc gets oxidized and act as the anode
Hydrogen gets reduced and acts as a cathode.
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 298 K
n = number of electrons in oxidation-reduction reaction = 2

The reaction involved in the question is as follows:-

[tex]Zn +2H^+--->Zn^{2+} +H_2[/tex].

The formula used in the question is as follows:-

[tex]E^o_{cell} = E^o_{cell} -\frac{2.303RT}{nF}log\frac{Zn^{2+}*pH_2}{(H^+)^2}[/tex]

After putting all the value,

[tex]E_{Cell}=0.76-\frac{2.303*(8.314)*(298)}{2*96500}log\frac{0.024*0.37}{(1.3)^2}}[/tex]

After solving the equation, the answer will be 0.69V.

Hence, the correct answer is 0.69V.

For more information, refer to the link:-

https://brainly.com/question/3523174