Calculate the potential of the indicator electrode:

Problem:

Calculate the potential of the indicator electrode at 25°C relative to S.C.E. at the equivalence point of the titration of 25.00 cm3  of 0.010 M Fe²⁺ with 0.010 MCr₂ O^2- ₇ in a sol. that is buffered at pH 3.50.

Answer:

On the addition of 25 cm³ of NaOH, the acid is completely neutralised giving NaCl.  The pH of the resulting solution is therefore 7 and E = 0.6560 V.

On additng 25.05 ml of NaOH, the excess volume of

NaOH = 0.05 cm³

Amount of NaOH in 0.05 cm³ solution = 0.05 × 0.1 = 0.005 mmol

Conc. of NaOH or OH¯ ions = 0.005 / 50.05 mol dm ^-3

Since [H+] [OH¯  ] = 10-14 at 25°C

[H+] =  (10^-14 × 50.05 )/0.005 mol dm -3

Hence pH = 10.0

The corresponding E is 0.833V.

5. The half-cell Potentials in this case are

Fe²⁺ ↔ Fe³⁺ +e^- ; E⁰Fe3+/Fe²⁺ =0.771 V(From Apeendix I)

C_r2O^2-₇ +14 H⁺ +6e^- ↔ 2Cr³⁺ +7H₂O; E⁰ _CrO^2-_7/Cr³⁺ =1.33 V(From Appendix I)

The overall Potential is

6 Fe²⁺ + Cr₂ O ^2-₇+ 14 H⁺     ↔ 6 Fe³⁺ + 2Cr³⁺ + 7H₂O;

The potential of the indicator electrode, relative to the S.H.E. is given by,

E_ln =0.771-0.0591 log {[Fe²⁺]}/{[Fe³⁺]}                    ...(1)

And by E_ln = 1.33 - 0.0591/6 log {[Cr³⁺]²/[Cr2O^2-₇][H⁺]¹⁴}   ...(2)

We must combine these two equations in such a manner as to eliminate the concentration of the reactants that cannot be calculated at the equivalence point.