Answer:
E = 2.02 V
Explanation:
In order to do this, we need to apply the Nernst equation which is:
E = E° - RT/nF lnQ
The value of RT/F can be simplified to just 0.059 because we are doing this experiment at 25 °C, and R and F are constants. so we need the value of Q which in this case is:
Q = [Mg²⁺] / [Ni²⁺]
We already have the concentrations, so, all we have left is the standard reduction potential, which are:
E° Mg = -2.38 V
E° Ni = -0.25 V
According to the overall reaction:
Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s)
we can see that one element is reducting and the other is oxidizing, so we need to write the semi equation of reduction for each element:
Mg(s) ---------> Mg²⁺ + 2e⁻ E° = 2.38 V oxidizing (Value of E° inverted)
Ni²⁺ + 2e⁻ -----------> Ni(s) E° = -0.25 V reducting
------------------------------------------------------------
Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s) E° = 2.13 V
We have the value of the standard potential, now we need to replace all given data into the nernst equation to solve for the cell potential:
E = 2.13 - 0.059/2 ln(0.757/0.0160)
E = 2.13 - 0.0295 ln(47.3125)
E = 2.13 - 0.11
E = 2.02 V
This is the cell potential
