Calculate the cell emf for the following reaction at 25^$\circ$C: 2Ag⁺(0.010 M) + H₂(1 atm) $\rarr$ 2Ag(s) + 2H⁺(pH = 10.0)

Question

Quizplus · Accepted Answer

To calculate the cell emf, we need to use the equation:

cell emf = E°cell - (0.0592/n) * log(Q)

Where:
E°cell = standard cell potential (given in tables)
n = number of electrons transferred in the balanced equation (here, n = 2)
Q = reaction quotient = [HS-]^2/[Ag+]^2

We can find E°cell for the reaction by subtracting the standard reduction potentials of the two half-reactions:

Ag+ + e- -> Ag (E°red = 0.80 V)
2H+ + 2e- -> H2 (E°red = 0.00 V)
2Ag+ + 2H+ + S2- -> 2Ag + H2S (E°cell = E°red,cathode - E°red,anode = 0.80 - 0.00 = 0.80 V)

Now we need to calculate Q. Since the pH is given as 10, we know that [H+] = 10^-10 M and [OH-] = 10^-4 M. Using the expression:

pH + pOH = 14

We can find that [HS-] = 10^-4 M. Additionally, since the pressure of H2S is given as 1 atm, we can assume that it is in its standard state and its concentration is 1 M.

Therefore, Q = [HS-]^2/[Ag+]^2 = (10^-4)^2/(0.010)^2 = 10^-8

Substituting these values into the equation for cell emf, we get:

cell emf = 0.80 - (0.0592/2) * log(10^-8) = 1.27 V

Therefore, the best choice is B.

Calculate the Cell Emf for the Following Reaction at 25 ∘\circ∘

Calculate the Cell Emf for the Following Reaction at 25^$\circ$