A concentration cell based on the following half reaction at 323K Ag+ + e- --> A

Question

A concentration cell based on the following half reaction at 323K

Ag+ + e- --> Ag SRP = 0.800V

has initial concentrations of 1.27M Ag+, 0.377 Ag+, and a potential of 0.03380V at these conditions. After 3.8hours, the new potential of the cell is found to be 0.01477V. What is the concentration of Ag+ at the cathode at this new potential?

The answer is 1.04M but i dont know how to approach to that answer.

A concentration cell based on the following half reaction at 323 K Ag^+ + e^- rightarrow Ag SRP = 0.800 V has initial concentrations of 1.27 M Ag^+, 0.377 M Ag^+, and a potential of 0.03380 V at these conditions. After 3.8 hours, the new potential of the cell is found to be 0.01477 V. What is the concentration of Ag^+ at the cathode at this new potential?

Explanation / Answer

For concentration cell Eo = 0 V

So,

E = -0.0592 log(cathode/anode)

let x be the change in concentration at equilibrium

0.03380 - 0.01477 = -0.0592 log(0.377 + x/1.27 - x)

0.983 = 1.477x

x = 0.665 M

So the concentration at the cathode at this new potential = 0.377 + 0.665 = 1.04 M

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