Calculate the emf of the following cell at standard conditions (temperature = 25o c, pressure = 1 atmosphere): ni | ni2+ | | cu2+ | cu if the concentrations of the two ions in solution are: (a) [ni2+] = 0.1 moles/mole and [cu2+] = 0.08 moles/mole of solution, and (b) [ni2+] = 0.4 moles/mole and [cu2+] = 0.3 moles/mole of solution.

a) +0.574 V

b) +0.573V

Explanation:

The EMF is the electromotive force, which is the property of a device that intends to generate electrical current. The EMF of a cell can be calculated by the Nernst equation:

Emf = E° - (0.0592/n)*logQ

Where E° is the standard reduction potential of the cell, n is the number of electrons involved in the reaction, and Q is the reaction quotient ([products]/[reactants]).

In the cell, a redox reaction happens. One substance oxides (lose electrons and become a cation), and the other one reduces (gains electrons and becomes an anion). Each reaction has a reduction potential (E), which indicates how easily is to the reduction happens (as higher E as easy).

For the overall reaction, E° = Ereduction - Eoxidation. To the cell given, Ni is oxidizing, and Cu⁺² is reducing, so, the half-reactions with its E (which can be found at tables), and the overall reaction are:

Ni(s) → Ni⁺² + 2e⁻ E = -0.24 V

Cu⁺² + 2e⁻ → Cu(s) E = +0.337 V

Ni(s) + Cu⁺² → Ni⁺² + Cu(s)

E° = +0.337 - (-0.24)

E° = +0.577 V

As we can see, there're 2 electrons involved in the reaction, so n =2.

The solids don't take place in the Q value, so:

Q = [Ni⁺²]/[Cu⁺²]

a) Q = 0.1/0.08 = 1.25

Emf = 0.577 - (0.0592/2)*log(1.25)

Emf = 0.577 - 0.003

Emf = +0.574 V

b) Q = 0.4/0.3 = 1.33

Emf = 0.577 - (0.0592/2)*log(1.33)

Emf = 0.577 - 0.004

Emf = +0.573 V

c. 75.7

i think that is the right