Answer:
I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)
Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)
a) E(Pt⁺²/Fe°) = - 1.668v
b) Process is Non-spontaneous if E(cell) < 0
Explanation:
Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔
Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)
As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.
E°(Fe⁺²) = -0.44v
E°(Pt⁺²) = +1.20v
E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)
= -0.44v - (+1.20v) = - 1.64v
[Fe⁺²] = 0.0066M
[Pt⁺²] = 0.057M
n = electrons transferred = 2
E(nonstd) = E°(std) - (0.0592/n)logQ);
Q = [Pt⁺²]/[Fe⁺²]
= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v
Also, if ΔG(cell) > 0 => indicates non-spontaneous process
ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)
