let's solve this step by step:
Given data:
E°(Mg^2+/Mg) = -2.36 V
E°(Ag^+/Ag) = 0.01 V
[Mg^2+] = 0.10 M
[Ag^+] = 0.0001 M
(i) To calculate the standard cell potential (E°cell), we use the formula:
E°cell = E°(cathode) - E°(anode)
For the given reaction, Mg is reduced (cathode) and Ag is oxidized (anode). Therefore:
E°cell = E°(Ag^+/Ag) - E°(Mg^2+/Mg)
= 0.01 V - (-2.36 V)
= 0.01 V + 2.36 V
= 2.37 V
(ii) Now, to calculate the cell potential (Ecell), we use the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
where n is the number of electrons transferred and Q is the reaction quotient.
In this case, n = 2 (since two electrons are involved), and Q is calculated using the concentrations of the reactants and products:
Q = [Ag^+]^2 / [Mg^2+]
Plugging in the values:
Q = (0.0001)^2 / 0.10
= 0.000001 / 0.10
= 0.00001
Now, substituting into the Nernst equation:
Ecell = 2.37 V - (0.0592/2) * log(0.00001)
= 2.37 V - (0.0296) * (-5)
= 2.37 V + 0.148 V
≈ 2.518 V
(iii) The cell reaction will be spontaneous if Ecell is positive. Since Ecell ≈ 2.518 V, which is positive, the above cell reaction will indeed be spontaneous.
