Question

The standard reduction potential for the Zn2+Ι Zn couple is -0.73 volts. What would the reduction potential be if the Zn2+ solution has a concentration of 0.08 M and is at a temperature of 15.9 °C? Enter your answer in volts, and use three significant figures. Include a "-" sign if your potential is negative, but do not use a "+" sign if it was positive.

Answer 1

Here E° = -0.73V

Zn⁺² (aq) +2e^-1 --------------->. Zn(s)

E​_cel = E° - (RT/nF)ln[zn⁺²]/[Zn]

R = gass constant = 8.314J/K-mol = 8.314VC/K-mol

T = temperature = 15.9°C = 15.9+273 = 288.9K

n = number of moles = 2

F = 96500C/mol

[Zn⁺²] = concentration of Zn ion = 0.08M

[Zn] = concentration of solid Zn = 1.0M ( concentration of a solid in solution always 1.0M)

Putting the all value

E_cel = (-0.73V) - [(8.314VC/K-mol)(288.9K)/2×(96500C/mol)]ln(0.08M)/(1.0M)

E_cel = (-.73V) - (0.01244V)(-2.5257)

E_{cel = -6985Volt}