Question

Calculate the cell potential, E, for the given reactions at 25.00 °C using the ion concentrations provided. Then, determine if the cells are spontaneous or nonspontaneous as written.

QUESTION 1)

Pt(s)+Fe2+(aq)−⇀↽−Pt2+(aq)+Fe(s)

[Fe2+]=0.0013 M[Pt2+]=0.048 M

E= ____ V?

The cell is not spontaneous or spontaneous.

QUESTION 2)

Cu(s)+2Ag+(aq)−⇀↽−Cu2+(aq)+2Ag(s)

[Cu2+]=0.017 M [Ag+]=0.017 M

E= ____ V?

The cell is not spontaneous or spontaneous.

QUESTION 3)

Co2+(aq)+Ti3+(aq)−⇀↽−Co3+(aq)+Ti2+(aq)

[Co2+] = 0.065 M [Co3+] = 0.025 M

[Ti3+] = 0.0060 M [Ti2+] = 0.0118 M

E= ____ V?

The cell is not spontaneous or spontaneous.

Answer 1

① The Nernst eam ix given by E E 0.0591 dog [a] A where E= EMF of cell E = standard EMF of cell. [Q] = Reaction Product Conc. quotient = Reactant conc. » Ptsd + Fe 2+ (aq) + P + 2 + + Fe(s) Q- [P+24] [Fe(s)] [Pt(3)] [Fe2+) [Fe(s)] = Pt(s) = 1 . Q = [Pt2t] [Fe247 ALSO PEU) 2 Pt² + cordiation), Ep +2+1 pt=1.200 A Fe 24 de Fe (Reduction), EF 24 F = -0.450 Reduction takes place at Cathode & oxidation at anode E EC-EA- – 0.450 1.200 = – 1.650 u from ean O ºla E = -465-0.0591. dog [P4243 2 [Fe21] E = -165-0.0591 loglo.one) E – 1.65-0.0463 E=-1169631) The cou be non spontaneous because the value of EMFI) -ve.

Tag+72 L ⓇT 1 culsd + 2 Agt upt + 2Ag (5) R = [wpt) { [uQ)= (190)-1] Anode - Cu 20curt (oxidation); Elu24 | M = 0.34 v Cathode-12 Agt det 20g (reduction), Eastlag = 0.BOV : E= E-EA = 0.60-6:34 = 0:46V E E -0.0591 log [cu 2+] 2 [Ag+] 2 = 0.46-0.0591 log 0.017 10.01772 = 0.46-0.0591x 7696 E = 0.407) 6 Rhy be spontaneous 11 colt & T: 3+ lost + T: 2+ Q = [0034] [T;) [co24] [Tist Anode corte 65t E63/63t = 1.82 v cathode Ti3t e ti?t . 37 1 777t = -0.37 V & E= E(-EA = -0.37-1.62 = -2.191 E-219-00591 dog (0.025 0.0148 ) (0.065* 0.0060 E- - 9-19 -0.0591(-0.1242) (E= - 2.18 V) NI

. Rh be non spontaneous.