A(voltaic) concentration cell is constructed using two zinc (zn) electodes and two aqueous solutions containing zn2+. if the concentrations of zn2+ in each of the two aqueous solutions are 1.25 m and 6.50 × 10-4 m, respectively, then the cell emf (ecell) is v.

a. −0.194
b. 0.194
c. −0.0972
d. 0.0972
e. 0.760

d. 0.0972

Explanation:

In a cell, a redox reaction happens, it means that one compound will reduce (gain electrons) and other will oxides (lose electrons). The reduction potential (E°) of a compound can be calculated by the redox between the compound and the hydrogen, which has E° = 0.00. As higher is the value of E°, as easy is for the compound to reduce. For a reaction, the value of E° = E° (the substance that reduces) - E°(the substance that oxides).

The cell emf can be calculated by the Nernst's equation:

Ecell = E° - (0.0592/n)*logQ

Where E° is the reduction potential of the cell (E°reduction - E°oxidation), n is the number of electrons being changing in the redox reaction, and Q is the reaction coefficient.

In this case, the cell is formed by the same compound, s E° = 0. Because Zn loses 2 electrons to form Zn⁺², n = 2, and the reaction must go from the highest concentration to the lowest, thus the reactant is where there's more Zn⁺², and the products where there's less Zn⁺²:

Q = [products]/[reactants]

Q = [Zn⁺²]less/[Zn⁺²]more

Ecell = -(0.0592/2)*log((6.50x10⁻⁴)/1.25)

Ecell = -0.0296*(-3.2840)

Ecell = 0.0972 V

the answer to your question is c