Aqueous iodide ion is oxidized to I_2(s) by Hg_2^2+(aq) I^-(aq) rightarrow I_2(s

Question

Aqueous iodide ion is oxidized to I_2(s) by Hg_2^2+(aq) I^-(aq) rightarrow I_2(s); Hg_2^2+ (aq) rightarrow Hg(l) a. write a balanced equation, b. calculate the standard emf, c. calculate Delta G degree at 298 K d. calculate the value of the equilibrium constant at 298 K a voltaic cell utilizes the following reaction and operates at 298 K: 3 Ce^4+(aq) + Cr(s) rightarrow 3 Ce^3+(aq) + Cr^3+(aq) a. What is the emf of this cell under standard conditions? b. What is the emf of this cell when [Ce^4+] = 3.0 M, [Ce^3+] = 0.10 M, and [Cr^3+] = 0.010 M? c. What is the emf of the cell when [Ce^4+] = 0.010 M, [Ce^3+] = 2.0 M, and [Cr^3+] = 1.5 M?

Explanation / Answer

Q5.

split

I- = I2

Hg2+2 = Hg

balance I and Hg

2I- = I2

Hg2+2 = 2Hg

balance e-

2I- = I2 + 2e-

2e- + Hg2+2 = 2Hg

add all

2I- + 2e- + Hg2+2 = 2Hg + I2 + 2e-

cancel common terms

2I- + Hg2+2 = 2Hg + I2

b)

calculate E°

I2(s) + 2 e 2 I +0.54

Hg22+ + 2 e 2 Hg(l) +0.80

invert I2

Hg22+ + 2 e 2 Hg(l) +0.80

2 I I2(s) + 2 e -0.54

E° = Ered + Eox = 0.80 + -0.54 = 0.26 V

c)

get

dG = -nF*E°cell

dG = -2*96500*0.26

dG = -50180 J/mol

dG = -50.180 kJ/mol

d)

find K

dG = -RT*ln(K)

K = exp(-dG/(RT))

K = exp(50180 /(8.314*298))

K = 625273675.951 = 6.25*10^8

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