The standard electrode potential for four half cells are as follows: A: Zn^2- (a

Question

The standard electrode potential for four half cells are as follows: A: Zn^2- (aq) + 2 e^- rightarrow Zn(s) E degree = -0.77 V B: Sr^2+ (aq) + 2e^- rightarrow Sr(s) E degree = -2.89 V C: Cu^2+ (aq) + 1 e^- rightarrow Cu^- (aq) E degree = +0.22 V D: Cr_2O^2-_7 (aq) + 14H^+ (aq) + 6e^- rightarrow 2Cr^3+ (aq) + 7H_2O(l) E degree = +1.33 V a) Which chemical species is the most powerful oxidizing agent? b) Calculate the standard cell potential, E degree_cell, for the oxidation of Zn by Cr_2O^2-_7 c) Write the Nernst equation for the cell in (b). a) How old is an earth rock found after an earth quake, containing a bearing mineral which had remained unaltered to the present time if 4.459 mg of found along with 15.39 mg of U-238?

Explanation / Answer

(14)

(a) Higher is the reduction potential higher is the ability to gain electrons and most is oxidizing nature.

SO, Cr2O72- is most powerful oxidising agent among the given.

(b)

E0cell = E0Cr2O72-/Cr3+ - E0Zn2+/Zn = + 1.33 - ( - 0.77 ) = + 2.10

(c)

Over all celll reaction is ,

Cr2O72- (aq.) + 3 Zn (s) + 14 H+ (aq.) -------------> 2 Cr3+ (aq.) + 3 Zn2+ (aq.) + 7 H2O (l)

Nernst equation,

Ecell = E0cell - (0.0591 / n) *Log[Cr3+]2[Zn2+]3 / [Cr2O72-][H+]14

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