Iodine (I2) is often used by backpackers and travelers for the treatment of drin

Question

Iodine (I2) is often used by backpackers and travelers for the treatment of drinking water. However, many people find the taste of iodine offensive. As a result, vitamin C, a.k.a. ascorbic acid (C6H8O6), tablets are sold as a “neutralizer,” that can be added after water treatment to remove the iodine taste and color. The products of the reaction are iodide ions and dehydroascorbic acid (C6H6O6). Zinc metal could be used instead of Vitamin C to remove the iodine. This fact can be established by placing galvanized (iron coated with zinc) tacks into a solution of iodine, and seeing that the reddish color of the solution disappears as some of the zinc coating on the tacks dissolves. a) Write the balanced redox equation for this reaction (iodine with zinc metal). b) What is the standard cell potential for this reaction at 25ºC (i.e., what voltage would be measured in a voltaic cell that employed this reaction under standard conditions)? c) What would the cell potential be if the iodide (I–) and Zn2+ concentrations were each 0.010 M, and that of the iodine was 0.080M? d) What is G for this reaction, under the conditions described in part c above? e) What is the equilibrium constant for this reaction at 25ºC?

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Explanation / Answer

a) Reaction : I2 + Zn --> ZnI2

b) The half reactions are: I2 + 2e- ---> 2I- E= 0.54V and Zn2+ + 2e- ---> Zn E= -0.7626V

This cell would have the potential = Ecell= Ecatode - Eanode = 0.54V- (-0.7626V) = 1.3026V

c) Using Nernst Equation, E=Eº + 0.05915/n log([OX]/[RED])

E1 = 0.54V + (0.05915/2)log(0.08M/0.01M) = 0.5667V

E2 = -0.7626V + (0.05915/2)log(0.01M) = -0.82175V

Ecell= 0.5667V - (-0.82175V) = 1.3885V

d) DG= -nFE = -2(96485C)(1.3885V) = -267.929kJ/mol

e) DG= -RTln(k); k= e^(DG/-RT) = e^(-267929J/mol/(-8.314J/molK)(298K)) = 9.2343x10^46

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