Predict the species that will be reduced first the following mixture of molten s

Question

Predict the species that will be reduced first the following mixture of molten salts undergoes electrolysis. FeBr_3(l) Znl_2(l) MgBr_2(l) A. Zn^2+ B. Mg^2+ C. Br^- D. Fe^3+ E. I^- Determine E degree cell for the reaction: 2 Ag^+ + Mg rightarrow 2 Ag + Mg^2+. The half reactions are: Mg^2+ (aq) + 2 e^rightarrow Mg(s) E degree = -2.356 V Ag^+(aq) + e^rightarrow Ag(s) E degree = 0.800 V A. + 0.756 V B. -0.756 V C. +3.156 V D. -1.556 V E. +1.556 V Determine K_eq at 25 degree C for the following reaction: Pb^2+ (aq) cu(s) Cu2 (aq)+Pb(s) The half reactions are: Pb^2+(aq) 2 e^Pb(s) E degree = -0.125 V Cu^2+(aq) + 2 e^- Cu(s) E degree = +0.337 V A. 2 times 10^69 B. 2 times 10^-15 C. 2 times 10^-16 D. 2 times 10^8 E. 6 times 10^-4

Explanation / Answer

(17) Higher is the reduction potential more is ability to undergo reduction.

The species that we have in the molten mixture are,

Fe3+, Br- , Zn2+ , I- and Mg2+

Among these ions only cations undergo reduction at cathode,

So (D) Fe3+ is reduced first as it has higher reduction potential.

(18)

E0cell = E0Ag+/Ag - E0Mg2+/Mg = 0.800 - ( - 2.356) = + 3.156 V

(c)

(19)

E0cell = - 0.125 - 0.337 = - 0.462 V

DeltaG0 = - n F E0cell = - 2 * 96500 * (-0.462) = 89166 J

And

deltaG0 = - R T lnK

89166 = - 8.314 * 298 * lnK

lnK = - 36.0

K = 2 * 10-16

(b)

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