For each of the following reactions, balance the chemical equation, calculate th

Question

For each of the following reactions, balance the chemical equation, calculate the emf, and calculate G° at 298 K. (Use the smallest possible coefficients for H2O(l), H+(aq), and HO-(aq). These may be zero.)

(a) Aqueous iodide ion is oxidized to I2(s) by Hg22+(aq). __I-(aq) + ___Hg22+(aq) + ___H+ ___I2(s) __Hg(l) + ___H2O(l)

_____V

_____kJ

(b) In basic solution Cr(OH)3(s) is oxidized to CrO42-(aq) by ClO-(aq). ___Cr(OH)3(s) + ___ClO-(aq) + ___OH-(aq) ___CrO42-(aq) ___Cl-(aq) + ____H2O(l) emf

____V

____kJ

(c) In acidic solution copper(I) ion is oxidized to copper(II) ion by nitrate ion. ___Cu+(aq) + ___NO3-(aq) + ___H+(aq) ___Cu2+(aq) ___NO(g) + __H2O(l) emf

____V

____kJ

Explanation / Answer

A) Hg22+ (aq) + 2 I- (aq) + OH- + H+ ----> 2 Hg (l) + I2 (s) + H2O

from standard emf table

(I) Hg22+(aq) + 2 e ---> 2 Hg (l) , E(I) = +0.80V
(II) I (s) + 2e ----> 2 I(aq) , E(II) = +0.54V

First reaction occurs reduction second oxidation.
Since both reaction produce the same number of electrons you can directly combine them to the overall reaction:

The emf of the overall reaction equals the sum of the standard potential
E° = E(reduction) - E(oxidation) = 0.80V - 0.54V = 0.26V

G and emf are related as
G° = n·F·E°

where
n is the number of electrons exchanged (here n =2)
F = 96485C/mol Faraday's constant

Hence:
G° = 2 x 96485C/mol x 0.26V
= 50172.2J/mol
= 50.17kJ/mol

b) 2 Cr(OH)3 + 3 ClO-1 + 4 (OH-) --> 2 CrO4-2 + 3 Cl- + 5 H2O  

Cr(OH)3(s) -----> 3 e- + CrO4(2-) E = +0.13 volts

ClO- + 2 e- Cl- E = +0.89 volts

Eo = 1.02 volts

dG = nFE = (6e-) (96,500) (1.02volts) = 590,580 J

dG = 591 kJ

c) Cu+(aq) + NO3-(aq) + 4H+(aq) ----> Cu2+(aq) + NO(g) + 2H2O(l)

Cu+ ----> Cu2+ +1e- E = -0.15 V

NO3- ----> NO E = 0.964 V

E = 0.964 v + 0.15 = 1.114 v

dG = nFE = (1e-) (96,500) (1.114 volts) = 107501 J

dG = 107.501 kJ

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