Calculate K at 25°C for the reactions given below.

(a) Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)

(b) 2H2O2 2 H2O + O2
Given:
H2O2 + 2 H+ + 2e- 2 H2O ° = 1.78 V
O2 + 2 H+ + 2e- H2O2 ° = 0.68 V

To calculate the equilibrium constant (K) at 25°C for the given reactions, we can use the Nernst equation which relates the standard electrode potential (E°), the temperature (T), the number of electrons transferred (n), and the equilibrium constant (K):

E = E° - (0.0592 V/n) * log(K)

(a) For the reaction: Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)

The standard electrode potential for the half-reaction Zn2+(aq) + 2e- Zn(s) is -0.763 V (from tables).
The standard electrode potential for the half-reaction 2Ag+(aq) + 2e- 2Ag(s) is 0.799 V (from tables).

To calculate the overall reaction potential:
E° = E°(Ag+) - E°(Zn2+)
E° = 0.799 V - (-0.763 V)
E° = 1.562 V

Now, we can use the given Nernst equation:
E = E° - (0.0592 V/n) * log(K)

At 25°C (298 K):
E = 1.562 V - (0.0592 V/2) * log(K)

Simplifying the equation, we have:
1.562 V = 0.0312 V * log(K)

Dividing both sides by 0.0312 V:
log(K) = 1.562 V / 0.0312 V
log(K) = 50

Taking the antilogarithm of both sides:
K = 10^50

Therefore, the equilibrium constant (K) for the reaction Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s) at 25°C is 10^50.

(b) For the reaction: 2H2O2 2 H2O + O2

From the given data:
H2O2 + 2 H+ + 2e- 2 H2O ° = 1.78 V
O2 + 2 H+ + 2e- H2O2 ° = 0.68 V

We need to use the Nernst equation separately for each half-reaction.

For the oxidation half-reaction: O2 + 2 H+ + 2e- H2O2

E° = 0.68 V

For the reduction half-reaction: H2O2 + 2 H+ + 2e- 2 H2O

E° = 1.78 V

To calculate the overall reaction potential:
E° = E°(reduction) - E°(oxidation)
E° = 1.78 V - 0.68 V
E° = 1.10 V

Now, we can use the given Nernst equation:
E = E° - (0.0592 V/n) * log(K)

At 25°C (298 K):
E = 1.10 V - (0.0592 V/2) * log(K)

Simplifying the equation, we have:
1.10 V = 0.0296 V * log(K)

Dividing both sides by 0.0296 V:
log(K) = 1.10 V / 0.0296 V
log(K) = 37.16

Taking the antilogarithm of both sides:
K = 10^37.16

Therefore, the equilibrium constant (K) for the reaction 2H2O2 2 H2O + O2 at 25°C is 10^37.16.

To calculate K at 25°C for the given reactions, we first need to determine the relationship between K and the standard cell potentials (°) of the half-reactions involved.

For reaction (a), we have:

Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)

To calculate K, we need the standard reduction potential for each half-reaction involved. The standard reduction potential (E°) can be found in tables of standard electrode potentials.

The reduction half-reactions for (a) are:

Zn2+(aq) + 2e- Zn(s) (1)

Ag+(aq) + e- Ag(s) (2)

The standard cell potential (°) for reaction (a) is the difference between the standard reduction potentials of the half-reactions:

°(a) = °(1) - °(2)

Now, let's find the values for °(1) and °(2).

For the reaction (b), we have:

2H2O2 2 H2O + O2

To calculate K, we need the standard reduction potential for each half-reaction involved. The reduction half-reactions for (b) are:

H2O2 + 2 H+ + 2e- 2 H2O (3)

O2 + 2 H+ + 2e- H2O2 (4)

The standard cell potential (°) for reaction (b) is the difference between the standard reduction potentials of the half-reactions:

°(b) = °(3) - °(4)

Now, let's find the values for °(3) and °(4).

Given that °(3) = 1.78 V and °(4) = 0.68 V, we can now calculate the values of °(a) and °(b).

°(a) = °(1) - °(2)

°(b) = °(3) - °(4)

Next, we use the Nernst equation to calculate the values of E(a) and E(b) at 25°C.

E(a) = °(a) - (0.0592 V / n)log10(Kc(a))

E(b) = °(b) - (0.0592 V / n)log10(Kc(b))

where n is the number of moles of electrons transferred in the balanced equation of the half-reaction.

After calculating the values of E(a) and E(b), we substitute them back into the Nernst equation to solve for the respective Kc(a) and Kc(b) values.

Kc(a) = 10^((°(a) - E(a)) / (0.0592 V / n))

Kc(b) = 10^((°(b) - E(b)) / (0.0592 V / n))

Now, by plugging in the values of °(a), E(a), n, °(b), and E(b) into the equations above, and calculating the values of Kc(a) and Kc(b), we can determine the values of K for reactions (a) and (b) at 25°C.