Calculation of Single Electrode Potential of Zn electrode in deci molar ZnSO4 solution

Given information:
Eº of Zn electrode = -0.762 volts

To calculate the single electrode potential of Zn electrode in deci molar ZnSO4 solution, we use the Nernst equation which relates the electrode potential of a half-cell to the concentrations of the species involved in the redox reaction.

Nernst equation:
E = Eº - (RT/nF)lnQ

Where,
E = electrode potential
Eº = standard electrode potential
R = gas constant
T = temperature in Kelvin
n = number of electrons transferred in the redox reaction
F = Faraday constant
Q = reaction quotient

We know that the redox reaction in the Zn electrode involves the dissolution of Zn metal and the formation of Zn2+ ions in solution.

The balanced equation for this reaction is:
Zn(s) → Zn2+(aq) + 2e-

The reaction quotient for this reaction is given by the expression:
Q = [Zn2+]

Where [Zn2+] is the concentration of Zn2+ ions in the solution.

Substituting the given values in the Nernst equation, we get:
E = -0.762 - (8.314×298/2×96485)ln([Zn2+]/1)

Simplifying the above equation, we get:
E = -0.762 - 0.0296ln[Zn2+]

Now, we need to find the concentration of Zn2+ ions in the deci molar ZnSO4 solution.

ZnSO4 dissociates in water to give Zn2+ and SO42- ions.
ZnSO4 → Zn2+ + SO42-

The concentration of Zn2+ ions in the solution is equal to the concentration of ZnSO4, which is 0.1 M (since the solution is deci molar).

Substituting this value in the above equation, we get:
E = -0.762 - 0.0296ln(0.1)

Simplifying the above equation, we get:
E = -0.762 - 0.0296(-2.3026)

Therefore, the single electrode potential of Zn electrode in deci molar ZnSO4 solution is:
E = -0.762 + 0.068 = -0.694 V

Hence, the correct option is 2) -0.671 V (closest to -0.694 V).