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

Answer : The electrode potential of the cell is, 0.69 V

Explanation :

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) :

Reduction reaction (cathode) :

The balanced cell reaction will be,

Given :

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

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

Using Nernst equation :

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

= electrode potential of the cell = ?

= emf of the cell = 0.76 V

= 0.37 atm

= 0.024 M

= 1.3 M

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

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

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