Answer:
a) +0.574 V
b) +0.573V
Step-by-step 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