Question

Use standard reduction potentials to calculate the equilibrium constant for the reaction: Pb2+(aq) + Cu(s) Pb(s) + Cu2+(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant: AGº for this reaction would bv _ than zero. greater less

Answer 1

As per the standard reduction potential

Reduction potentials are

E°Pb2+⁄Pb = -0.126

E° Cu2+ ⁄Cu = 0.337

Now we will find the E° of the cell.

РА** +20 — Pь 6°C -0.126 Cu2+ + 2€ u E = 0.337. - As t° for cuttle is greath than E foL Pb2+ | Pb. - so , cu²t will easily reduce and Pb2t will not able to reduce easily. 1. So, here in this cell there will be reduction at coppel cul While oxidation at lead (Pb): Therefore, we will reverse the sign of E° (Pb2+1 Pb) to make it Epbipb2t (as there is oxidation possible). Pb Pb2+ the E = 0.126 E of cell is EPb/Pb2+ + Ecutlu on putting iting the values we get the v 0.126 +0.337 SO, E = 0.463

So E° of the cell is 0.463

Also the formula for determining equilibrium constant using nernst equation is

Log k = nFE°⁄2.303RT…….equation 1

Where

k=? (equilibrium constant, whose value need to be determined)

n=2   (number of electrons involved in the reaction)

F = 96500 (Faraday constant)

E°= 0.463 (as we have calculated)

R= 8.314 (Gas constant)

T= 298K (we took room temperature because no specific temperature is given)

As the values of T, F, and R are constant so we can put these values and rearrange equation 1 and the formula becomes

Log k = nE°⁄0.0591…equation 2

Now we put the values of n and E° in the above equation

Log k =(2×0.463)⁄0.0591

          = (2×0.463)⁄0.0591

           =15.667

On taking antilog

k=4.6451×1015

So, the equilibrium constant is 4.6451×1015

As ∆G°= -nFE°

E° = 0.463 (We have already calculated the value of E°)

n = 2

F = 96500

On putting the values we get

∆G°= -2×0.463 × 96500

The value ∆G°= -89359

The value of ∆G° is negative, so the value of ∆G° is less than 0.