Answer:
[Ni²⁺] = 1.33 M
Explanation:
To do this, we need to use the Nernst equation which (in standard conditions of temperature)
E = E° - RT/nF lnQ
However R and F are constant, and the reaction is taking place in 25 °C so we can assume the nernst equation like this:
E = E° - 0.05916/n logQ
As the nickel is in the cathode, this means that this element is being reducted while Cadmium is being oxidized, therefore the REDOX reaction would be:
Cd(s) + Ni²⁺(aq) --------> Cd²⁺(aq) + Ni(s)
With this, Q:
Q = [Cd²⁺] / [Ni²⁺]
Now, we need to know the value of the standard reduction potentials, which can be calculated with the semi equations of reduction and oxidation:
Cd(s) ------------> Cd²⁺ + 2e⁻ E°₁ = 0.40 V
Ni²⁺ + 2e⁻ -------------> Ni(s) E°₂ = -0.25 V
E° = E°₁ + E°₂
E° = 0.40 - 0.25 = 0.15 V
Now that we have all the data, we can solve for the [Ni²⁺]:
0.133 = 0.15 - 0.05916/2 log(5/[Ni²⁺])
0.133 - 0.15 = -0.05916/2 log(5/[Ni²⁺])
-0.017 = -0.02958 log(5/[Ni²⁺])
-0.017/-0.02958 = log(5/[Ni²⁺])
0.5747 = log(5/[Ni²⁺])
10^(0.5747) = 5/[Ni²⁺]
[Ni²⁺] = 5/3.7558
<h2>
[Ni²⁺] = 1.33 M</h2>
