Question

You are titrating 100.0 mL of 0.0400 M Fe2+ in 1 M HCIO4 with 0.100 M Ce4+ to give Fe3+ and Ce3+ using Pt and calomel electrodes to find the endpoint. (a) Write the balanced titration reaction. > (b) Complete the two half reactions for the Pt electrode. Ce té E° = 1.70 V Fe +e = E° = 0.767 V [Fe2+11 1 - 0.241 E= 0.767 – 0.05916) (c) From the list in the column at the right, select the two correct Nernst equations for the cell voltage. (Each applying at different points in the titration.) Eº of the calomel electrode is 0.241 V. E= 0.767 -0.05916 log -0.241 (A) De+1 -0.241 (c) E= 1.70 – 0.05916 log -1-0.241 SIDDDDD @ o ê a e o (D) E= 1.70 – 0.05916loge -0.241 = 1.70 – 0.059161 il-0.241

Answer 1

(a) Balanced titration reaction: Fe²⁺ + Ce⁴⁺ ---> Fe³⁺ +Ce³⁺---------------(1)

(b) Half reactions: Ce⁴⁺ + e^- ---> Ce³⁺ E⁰=1.70 V     -----------(1.1)

                            Fe³⁺ + e^- ---> Fe²⁺   E⁰=0.767 V -----------(1.2)

(c) Nernst equation for the reaction:

               A⁺+B->A+B⁺.

E_cell=(E⁰_A⁺/A-E⁰_B⁺/B) - 0.05916 log[[B⁺][A]/[B][A⁺]] ----------(2)

Here E⁰_A⁺/A > E⁰_B⁺/B

and E⁰_B⁺/B is taken as 0.241 V.

So using Eq. (2), one gets

E_cell=0.767-0.241 - 0.05916 log[[Fe²⁺]/[Fe³⁺]]

E_cell=1.70-0.241 - 0.05916 log[[Ce³⁺]/[Ce⁴⁺]].

So the correct answers are (A) and (G).

(d) Before equivalance point:

Nernst equation for the half-reaction (Eq. 1.2):

E_cell=0.767 - 0.05916 log[[Fe²⁺]/[Fe³⁺]] -----------(3)

Suppose, V ml of Ce⁴⁺ is added.

Moles of Fe²⁺ present= 100*0.04=4 mM.

After reaction with V*0.1 mM Ce⁴⁺, [Fe²⁺]=4-V*0.1 mM

Number of Fe³⁺ produced=V*0.1 mM

So from Eq. (3), one has

E=0.767 - 0.05916 log[(4-V*0.1)/(V*0.1)] ----------- (4)

Using Eq. (4), the following result can be obtained:

V=1 ml         E= 0.673 V

V=20 ml      E= 0.767 V

V=38 ml      E= 0.843 V

-------------------------------------------------

V=40 ml      equivalence point

                   E=(1.7+0.767)/2=1.23 V

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After equivalance point:

E_cell=1.7 - 0.05916 log[[Ce³⁺]/[Ce⁴⁺]] -----------(3)

Initial moles of Fe²⁺=moles of Ce³⁺ formed=4 mM

moles of Ce⁴⁺=V*0.1- 4.0

E_cell=1.7 - 0.05916 log[4/(V*0.1- 4.0)] -----------(3)

V=45 ml      E= 1.65 V

V=80 ml      E= 1.70 V