Question

Ascorbic asic (0.0100 M) was added to 10.0 mL of 0.0200 M Fe3+ at pH = 0.30, and the potential was monitored with Pt and saturated Ag/AgCl electrodes.

Dehydroascorbic acid + 2H⁺ + 2e^- --> ascorbic acid + H₂O (E^o = 0.390 V)

Using E^o = 0.767 V for the Fe³⁺/Fe²⁺ couple, calculate the cell voltage when 5.0, 10.0, and 15.0 mL of ascorbic acid have been added.

Correct answers are: 0.570 V, 0.307 V, and 0.184 V but not sure how to get there

Answer 1

Given data

Fe3+ + e-     Fe2+      Eo =0.767 V

(a) When 5.0 ml of ascorbic acid of 0.01 M was added

No. of moles of ascorbic acid = 0.01 M x 5 mL = 0.05 mM

No. of moles of Fe3+ = 0.02M x 10 mL = 0.2 mM

No. of moles of Fe2+ formed = 0.05 mM/ 15 mL = 0.0033 M

No. of moles of Fe3+ remained = 0.2 mM -0.05mM/15mL =0.15 mM/15mL =0.01 M

We need to subtract potential of ascorbic acid (0.390V/2 = 0.195 V)

From the Nernst equation

E cell = EoCell – (0.0591 V/n) log(Fe2+/ Fe3+) – 0.195 V

E = 0.767-(0.0591/1 log( 0.0033/0.01)- 0.195 V

    = 0.767- (0.0591 x -0.4814)-0.195 V

    = 0.767 +0.028 = 0.795-0.195 V = 0.6 V

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

No. of moles of ascorbic acid = 0.01 M x 10 mL = 0.1 mM

No. of moles of Fe3+ = 0.02M x 10 mL = 0.2 mM

No. of moles of Fe2+ formed = 0.1 mM/ 20 mL = 0.005 M

No. of moles of Fe3+ remained = 0.2 mM -0.1mM/15mL =0.1 mM/20mL =0.005 M

E cell = EoCell – (0.0591 V/n) log(Fe2+/ Fe3+) – 0.195 V

As concentrations of Fe2+ and Fe3+ are same, log 1 =0

E cell = EoCell – 0.197 V = 0.767-0.197 = 0.570 V

(c) When 15.0 ml of ascorbic acid of 0.01 M was added

No. of moles of ascorbic acid = 0.01 M x 15 mL = 0.15 mM

No. of moles of Fe3+ = 0.02M x 10 mL = 0.2 mM

No. of moles of Fe2+ formed = 0.15 mM/ 25 mL = 0.006 M

No. of moles of Fe3+ remained = 0.2 mM -0.15 mM/25mL =0.05 mM/25mL =0.002 M

E cell = EoCell – (0.0591 V/n) log(Fe2+/ Fe3+) – 0.197 V

        = 0.767 -0.0591/1 log(0.006/0.002) -0.197V

         = 0.767 -0.0282-0.197 V = 0.542 V