What is K_eqat 25 °C of the spontaneous cell made from Ag⁺(aq)/Ag(s)(0.80 V)and Cl₂(g)/Cl^-(aq)(1.36 V)half cells? A)9 × 10^-74 B)1 × 10^-19 C)1 × 10⁷³ D)9 × 10¹⁸ E)1 × 10²⁷

Question

Quizplus · Accepted Answer

The cell potential can be calculated using the formula: Ecell = Ecathode - Eanode.

Ecathode = Standard reduction potential of Ag+ to Ag(s) = 0.80 V 
Eanode = Standard reduction potential of Cl2(g) to 2Cl- = -1.36 V

Ecell = 0.80 - (-1.36) = 2.16 V

Next, we can use the Nernst equation to calculate the cell potential at 25 °C (298 K).

Ecell = E°cell - (RT/nF)lnQ

E°cell = 2.16 V (from above) 
R = 8.314 J/mol-K 
n = number of moles of electrons transferred = 2 (from the balanced equation) 
F = Faraday's constant = 96,485 C/mol 
Q = reaction quotient = [Ag+] / [Cl2]

At equilibrium, Q = K, the equilibrium constant. Since the cell reaction is spontaneous, K > 1.

We can assume that [Ag+] is negligible compared to [Cl2] since it is a solid metal and its concentration does not change appreciably. Therefore:

Q ≈ [Cl2]

We don't know the concentration of Cl2 in the gas phase, but we can use the ideal gas law to relate it to the pressure:

PV = nRT

P = pressure of Cl2 
V = volume of Cl2 (not given but cancels out) 
n = number of moles of Cl2 = 1 from the balanced equation (since 1 mole of Cl2 produces 2 moles of electrons transferred) 
T = temperature in Kelvin = 298 K

P = nRT/V = (1 mol)(8.314 J/mol-K)(298 K) / (1 atm * 101325 Pa/atm) = 0.024 atm

Now we can calculate Q:

Q = [Cl2] = P/Keq = 0.024 atm / K

Substitute into the Nernst equation:

Ecell = 2.16 V - (8.314 J/mol-K * 298 K / (2 * 96,485 C/mol)) * ln(P/Keq)

Solving for ln(P/Keq):

ln(P/Keq) = (2.16 V - Ecell) * (2 * 96,485 C/mol) / (8.314 J/mol-K * 298 K)

ln(P/Keq) = 35.77

P/Keq = e^(35.77)

P/Keq = 9.1 x 10^15

Since Q = P/Keq:

Q = 9.1 x 10^15

The answer is D, 9 × 10^118.

What Is Keq at 25 °C of the Spontaneous Cell

What Is K_eqat 25 °C of the Spontaneous Cell