7. A procedure called coulogravimetric analysis combines a weight determination

Question

7. A procedure called coulogravimetric analysis combines a weight determination with a measurement of the quantity of electricity in a controlled-potential electrolysis to analyze bromide chloride mixtures. An aqueous sample containing Br and Cl in exactly 25.00 mL of solution was oxidized at a silver anode to form both AgBr and AgCl. The mass (weight) of the silver anode before the electrolysis was 7.1864 grams, and the mass (weight) of the silver anode after the electrolysis was 7.4612 grams. In addition, the electrolysis involved the passage of 435.7 coulombs of electricity. Calculate the molar concentrations of Br and Cl in the original 25.00-mL sample.

Explanation / Answer

7. According to the publication by McIver (Coulogravimetric Determination of Halides; W. M. MacNevin, B. B. Baker, and R. D. Mclver; Anal. Chem., 1953, 25 (2), pp 274–277),

Increase in the weight of anode = weight of chloride + weight of Bromide ------------- (eq.1)

and

(Coulombs required/96500) = wt. of chloride/35.5 + wt of bromide/79.92 -----------------(eq. 2)

Now the initial weight of anode = 7.1864g

Final weight of anode = 7.4612g

Thus increase in the weight of anode = 7.4612-7.1864 = 0.2748 g

Since there is no mention about the left over Cl- and Br- after the eletrolysis, we can assume that the whole Cl- and Br- were deposited on the anode.

Lets take

Weight of Cl- deposited = X and

Weight of Br- deposited = Y;

Therefore X + Y = 0.2748 ------------ (eq. 3)

And

435.6/96500 = X/35.5 + Y/79.92 -------------(eq.4);

From eq.3 we get Y = 0.2748-X -----------(eq. 5)

Replacing this value for Y in eq. 4, we get

435.6/96500 = X/35.5 + [(0.2748-X)/79.92] ---------- (eq. 6);

Solving eqn. 6 for X we get X= 0.07;

Replacing X in eq.3 and solving for Y we get

Y = 0.2045

Therefore, concentration of Cl- in the original solution = (0.07*1000)/(25*35.5) = 0.0788M

And similarly, concentration of Br- in the original solution = (0.2045*1000)/(25*79.92) = 0.102M;

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