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If the E of the following cell is 0.945 V, what is the pH in the anode compartment? cell Pt(s)∣H₂​(1.00 atm)∣H+(aq)∥Ag⁺(0.100M)∣Ag(s)

User TechnoCore
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The pH in the anode compartment is -1.459.

Given the cell:

Pt(s) | H₂(1.00 atm) | H+(aq) || Ag⁺(0.100M) | Ag(s)

And the following information:

E_cell = 0.945 V

P_H2 = 1.00 atm

M_Ag = 0.100 M

The pH in the anode compartment can be calculated using the Nernst equation and the following steps:

Calculate the standard cell potential (E°).

Calculate the reaction quotient (Q).

Calculate the pH.

The Nernst equation is:

E = E° - (RT ln(Q)) / (nF)

Where:

E is the cell potential (V)

E° is the standard cell potential (V)

R is the gas constant (J/mol·K)

T is the temperature (K)

ln is the natural logarithm

Q is the reaction quotient

n is the number of electrons transferred

F is Faraday's constant (C/mol)

The reaction quotient is:

Q = [Ag⁺] / (P_H2)^(1/2)

The number of electrons transferred is 1.

Plugging in the values, we get:

0.945 = 0.7996 - (8.314 * 298 * ln(Q)) / (96485 * 1)

Solving for Q, we get:

Q = 1.729 x 10^(-3)

The pH is:

pH = -log10(Q * M_Ag / P_H2^(1/2))

Plugging in the values, we get:

pH = -1.459

Therefore, the pH in the anode compartment is -1.459.

User Flauwekeul
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