Question

Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine if the cells are spontaneous or nonspontaneous as written. Standard reduction potentials (E°red) may be found here.'1) Pt(s)+Fe2+(aq)\rightleftharpoonsPt2+ (aq)+Fe(s)[Fe2+]=0.0066M [Pt2+]=0.057ME= ? V2) Cu(s)+2Ag+(aq)\rightleftharpoonsCu2+(aq)+2Ag(s)[Cu2+]=0.013M [Ag+]=0.013ME= ?V3) Co2+(aq)+Ti3+(aq)\rightleftharpoonsCo3+(aq)+Ti2+(aq)[Co2+]=0.050M [Co3+]=0.030M[Ti3+]=0.0055M [Ti2+]=0.0110ME=?VCalculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.796 M and [Sn2 ] = 0.0170 M. Standard reduction potentials can be found here.Mg(s)+Sn2+(aq)\rightleftharpoonsMg2+(aq)+Sn(s)E=?V

Answer 1

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)