Based on the following electrochemical cell, what is the standard reduction potential of metal M at 298 K?

Question

Quizplus · Accepted Answer

To find the standard reduction potential of metal M, we need to look at the half-reaction where M is being reduced. 
From the given cell, the reduction half-reaction is:
M+(aq) + 2e- -> M(s)    E°cell = +0.80 V
Using the Nernst equation, we can calculate the standard reduction potential of M:
E° = E°cell - (0.0257 V/K) * ln(K)
where K is the equilibrium constant of the reaction, which can be found using the concentrations of the species:
K = [M]/([M+]^1)
where [M] represents the concentration of the solid metal, and [M+] represents the concentration of the M+ ion in solution.
Since the metal M is in its standard state, its concentration is 1 M, and since it is being oxidized in the anode, its concentration is increasing. Thus, [M+] is decreasing, and at equilibrium, [M] = [M+]. So we can simplify the equilibrium constant:
K = [M] / [M+]  = 1
Plugging this into the Nernst equation, we get:
E° = E°cell - (0.0257 V/K) * ln(1)  =  E°cell = +0.80 V
Therefore, the standard reduction potential of metal M is +0.80 V.

Based on the Following Electrochemical Cell, What Is the Standard