# REDOX TITRATIONS PART-2: Curve, Equations, Summary and MCQ for GPAT, GATE and CSIR NET JRF

## REDOX TITRATIONS PART-2: Curve, Equations, Summary and MCQ for GPAT, GATE and CSIR NET JRF

### REDOX TITRATION CURVE :-

• Redox titration is monitored by observing the change of a electrode potential.
• The titration curve is a drawn by taking the value of this potential (E) vs the volume of  the titrant added.
• The redox reaction is rapid and the system is always in equilibrium throughout the titration.
• The electrode potential of the two half reaction are always identical.
• If we  consider the oxidation of fe(II) with standard ce(IV) than we can we can write the equation as follows :-

Fe2+ +Ce4  ⇔  Fe3 +Ce3

• The number of electron transferred will be equal to 1.
• The electrode potential of the two half reaction will be always identical.
• For iron the electrode potential will be

E = EFe +0.059/1 Log [Fe2]/[Fe3]

• For cerium the electrode potential will be

E= Ece +0.059/1 Log [Ce3]/[Ce4]

• The utilization of the either equation is based on the stage of titration is based on the stage of titration prior to the equivalence point the conc. of Fe(II) & Fe (II) are appreciable compare to Ce (IV) ion which is negliglible because of the presence of large excess of Fe(II)
• Beyond the equivalence point the conc. of Ce(IV) & Ce(III) is readily computed from the addition and the electrode potential for the Ce (IV) could be used.
• At equivalence point the conc. of the oxidized and reduced forms of the two species are such that their attraction for electrons are identical.
• At this point the reactant species conc. & product species conc. ratios are known and they are utilized to calculate the potential at this point is called as equivalence point potential.

∗ 50 ml of 0.005m Fe² is titrated with 0.1 M Cein a sulphuric acid media at all times calculate the potential of the inert electrode in the solution at various intervals in the titration & plot the titration curve use 0.68V as the format potential of the Fe2+ –  Fesystem in H2SO4 & 1.44V for the CeCe3system.

Fe2+ +Ce4  ⇔  Fe3 +Ce3

Initial Step :- After addition of 5.0 ml of Ce4

As because the is too small we are considering the iron electrode potential to calculate the solution potential.

[Fe³] = 5 ml × 0.10 M (50+5)ml

= 0.5 m mol/55 ml

= 0.00909

Similarly [Fe²] = (50 ml × 0.05 M -5 ml ×0.1M )/55 ml

= 2 mmol / 55 ml =0.0363

substituting the value to the standard electrode potential equation we can get

E=Eº+0.059 log [Fe²]/[Fe³]

E=Eº + 0.059 log [0.0363/0.00909]

E = 0.68+0.036 = 0.716V

Step 2 : Equivalence point :-

At equivalence point in the titration of Fe(II) and Ce (IV) the potential of the solution is controlled by both the half reaction.

E = E + 0.059 log [Ce3+]/[Ce4+]

E = E +0.059 log [Fe²]/[Fe3+]

Step 3 After addition of 25.1 ml Ce

At this stage the concentration of Fe(II) negligible hence we will utilise the Ce (IV) electrode potential to calculate the solution potential.

[Ce4+]= (0.1ml×0.1M)/(50+25.1)ml

= 0.01 mmol / 75.1 ml =0.0001

[Ce3+] = (25ml × 0.1 M)/(50+25.1)ml

= 0.0332

E = Eº – 0.059 log [Ce3+]/[Ce4+]

= 1.44 +0.059 log 0.0332/0.0001

= 1.44 +0.059 log332

= 1.44 +0.148

= 1.58V

SUMMARY :-

• The greater the difference in reduction potential between analyte and titrant the sharper will be the end point
• The voltage at any point in this titration depends only on the ratio of reactant it will be independent of dilution.
• Prior to the equivalence point the half reaction involving analyte is used to find the voltage because the conc. of both the oxidized and the reduced forms of analyte are known.
• After the equivalence point the half reaction involving titrant is employed.
• At the equivalence point the half reaction involving titrant is employed.
• At the equivalence point both half reaction are used simultaneously to find the voltage.

# MCQ

1. What is the ratio of  Sn+2/Sn+4

If Ecell= +0.13 and E Sn+2 /Sn+4 = +0.15

a. 4.02 × 109

b.1.06 × 109

c. 2.988 × 109

d. 0.56 ×109

2. When ECell  of SCE and quinhydron electrode is 0.321 V then what is the pH ?

a.2.25

b. 1.25

c. 0.23

d.3.25

3.Find emf of the following cell at 25 c Cr/Cr+3 (0.1) // Fe+2 (0.01M)/Fe

E Cr/Cr+3 = -0.74 E Fe+2/Fe = -0.44 V

a.0.30V

b. 0.26 V

c.0.11 V

d.0.59 V

4. If a Mg (S)/ Mg2+(aq)//Fe2+(aq)/Fe (s)

Mg+2=-2.37V

Fe+2 = -0.45V

So calculate Ecell ?

a.+3.2V

b.-0.92V

c.+1.92V

d.-0.1v

1.C

2.A

3.B

4.C

REFERENCE :-

PHARMACEUTICAL ANALYSIS THIRD EDITION OF DR. RAVI SANKAR (PG NO. 20.1-20.7).

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