Ascorbic acid (0.0100 M) was added to 10.0 mL of 0.0200 M Fe3+ in a solution buf

Question

Ascorbic acid (0.0100 M) was added to 10.0 mL of 0.0200 M Fe3+ in a solution buffered to pH 0.30, and the potential was monitored with Pt and saturated Ag | AgCl electrodes. dehydroascorbic acid + 2 H+ + 2 e equilibrium reaction arrow ascorbic acid + H2O E = 0.390 V

(a) Write a balanced equation for the titration reaction. (Use the lowest possible coefficients. Omit states-of-matter. Use "ascorbic acid" and "dehydroascorbic acid".)

(b) Using E = 0.767 V for the Fe3+ | Fe2+ couple, calculate the cell voltage when 4.9, 10.0, and 15.3 mL of ascorbic acid have been added. (Hint: Refer to the calculations here.)

Explanation / Answer

(a) Balanced equation for titration,

2Fe3+ + ascorbic acid + H2O ---> 2Fe2+ + dehydroascorbic acid + 2H+

(b) Fe3+ + e- ---> Fe2+ Eo = 0.767 V

(i) When 4.9 ml of 0.01 M ascorbic acid was added

moles of ascorbic acid = 0.01 M x 4.9 ml = 0.049 mmol

moles of Fe3+ = 0.02 M x 10 ml = 0.2 mmol

formed [Fe2+] = 0.049 mmol/14.9 ml = 0.0033 M

remaining [Fe3+] = 0.151 mmol/14.9 ml = 0.010 M

E = [0.767 - 0.0592 log(0.0033/0.010)] - 0.197 = 0.601 V

(ii) When 10.0 ml of 0.01 M ascorbic acid was added

moles of ascorbic acid = 0.01 M x 10 ml = 0.1 mmol

moles of Fe3+ = 0.02 M x 10 ml = 0.2 mmol

This is half equivalence point

E = 0.767 - 0.197 = 0.570 V

(iii) When 15.3 ml of 0.01 M ascorbic acid was added

moles of ascorbic acid = 0.01 M x 15.3 ml = 0.153 mmol

moles of Fe3+ = 0.02 M x 10 ml = 0.2 mmol

formed [Fe2+] = 0.153 mmol/25.3 ml = 0.006 M

remaining [Fe3+] = 0.047 mmol/25.3 ml = 0.002 M

E = [0.767 - 0.0592 log(0.006/0.002)] - 0.197 = 0.542 V

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