Using standard electrode potentials calculate δg∘rxn and use its value to estimate the equilibrium constant for each of the reactions at 25 ∘c.

The value of K for the equation Pb2+(aq)+Mg(s)->Pb(s) + Mg2+ (aq) is K = 2.35 X10^75 The value of k for the equation Br2 (l)+2Cl-(aq)->2Br-(aq)+Cl2 (g)  is

          K = 1.26 X10^-10

The value of k for the equation MnO2 (s) +4H+ (aq) + Cu (s) -> Mn2+ (aq) +2H2O(l) + Cu2+(aq) Eo is K = 1.17 X10^30

Explanation:

We have the formula,

By the Nernst equation:

E = Eo - (0.0592/n)(log Q)

Put E=0which at equilibrium becomes

0 = Eo - (0.0592/n) (log K)

which re-write the equation as

log K = Eo / (0.0592/n)

1.Pb2+(aq)+Mg(s)->Pb(s) + Mg2+ (aq)

Finding the value of Eo we get,

Pb2+(aq)--> Pb(s) Eo = –0.125

Mg(s) -> Mg2+ (aq) Eo = + 2.356

Pb2+(aq)+Mg(s)->Pb(s) + Mg2+ (aq) Eo = 2.231

Subsitute in this formula we get

log K = Eo / (0.0592/n)

log K = 2.231 / (0.0592/2)

log K = 75.37

K = 2.35 X10^75

The value of K for the equation Pb2+(aq)+Mg(s)->Pb(s) + Mg2+ (aq) is K = 2.35 X10^75

2.Br2 (l)+2Cl-(aq)->2Br-(aq)+Cl2 (g) Eo = -0.293

Similarly, for the other equation  follow the same procedure

The equation is written in the form of non-spontaneous

2Cl-(aq) -> Cl2 Eo = -1.358

Br2 (l) --> 2Br-(aq Eo = 1.065

Br2 (l)+2Cl-(aq)->2Br-(aq)+Cl2 (g) Eo = -0.293

log K = -0.293 / (0.0592/2)

log K = - 9.90

K = 1.26 X10^-10

The value of k for the equation Br2 (l)+2Cl-(aq)->2Br-(aq)+Cl2 (g)  is

K = 1.26 X10^-10

3.MnO2 (s) +4H+ (aq) + Cu (s) -> Mn2+ (aq) +2H2O(l) + Cu2+(aq)

Similarly, for the other equation  follow the same procedure

The equation is written in the form of non-spontaneous

MnO2 (s) -> Mn2+ Eo = 1.23

Cu (s) -> Cu2+ Eo = - 0.340

MnO2 (s) +4H+ (aq) + Cu (s) -> Mn2+ (aq) +2H2O(l) + Cu2+(aq) Eo = 0.89

log K = 0.89 / (0.0592/2)

log K = 30.067

K = 1.17 X10^30

The value of k for the equation MnO2 (s) +4H+ (aq) + Cu (s) -> Mn2+ (aq) +2H2O(l) + Cu2+(aq) Eo is K = 1.17 X10^30

High-energy and unstable.

The answer is b because a,c,and d are opinons