Question

A concentration cell is constructed using two metal (M) electrodes with M2+ concentrations of 0.10 M and 1.00 × 10–5 M in the two half-cells. Determine the reduction potential of M2+ given that the potential of the cell at 25°C is

a. 0.118 V.
b. Cannot be determined with the information given.
c.–0.118 V
d. 0 V
e. +0.118 V
f. None of these choices are correct.

Answer 1

The answer is: d. 0 V

Step-by-step explanation:

The reactions are:

M⁺² + 2e⁻ = M

M = M⁺² + 2e⁻

According the Nernst expression:

`E_(cell) =E_(o,cell) -(0.059)/(n) log([Red])/([O*d])`

Where

n = 2

Ecell = 0.118 V

[Red] = 1x10⁻⁵M

[O*d] = 0.1 M

Clearing Eo,cell:

`E_(o,cell) =E_(cell) +(0.059)/(n) log([Red])/([O*d])=0.118+(0.059)/(2) log(1x10^(-5) )/(0.1) =0.118-0.118=0V`