Calculate the cell emf for the following reaction at 25^$\circ$C: 2Ag⁺(0.010 M) + H₂(1 atm) $\rarr$ 2Ag(s) + 2H⁺(pH = 6.0)

Question

Quizplus · Accepted Answer

First, we need to write the half-reactions and determine their standard reduction potentials at 25°C:

Ag+ + e- → Ag(s)         E°red = +0.80 V 
H+ + e- → 1/2H2(g)     E°red = 0.00 V 
H2S(g) + 2e- → 2HS-     E°red = -0.17 V

Then, we can calculate the cell emf using the Nernst equation:

Ecell = E°cell - (RT/nF)ln(Q)

where E°cell is the standard emf, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred, F is the Faraday constant, and Q is the reaction quotient.

The balanced equation for the reaction is:

2Ag(s) + 2HS- + S2O32- → 2Ag2S(s) + SO42- + H2O

So, the cell emf is:

Ecell = E°cell - (RT/nF)ln(Q)
Ecell = (E°red,cathode - E°red,anode) - (RT/nF)ln(Q)
Ecell = (0.00 - 0.80 - (-0.17)) - (8.3145 J/mol*K)(298 K)/(2 mol)(96485 C/mol)ln[(1/0.010)(1/atm)]
Ecell = 1.04 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^$\circ$