Question

Consider the titration of 50.0 mL of 0.0500 M Cu+ with 0.1000 M Fe3+ to give Cu2+ and Fe2+ using Pt and standard H+/H2 reference electrode to find the end point. The standard electrode potentials are E0Cu+/Cu2+=0.161 V and E0Fe2+/Fe3+=0.767 V. Calculate the actual electrode potential, E, at the following volumes of Cu2+ (i) 15, (ii) 25, and (iii) 26 mL

Answer 1

Given data

The titration reaction

Cu+ + Fe3+ = Cu2+ + Fe2+

The reduction reactions are

Fe3+ + 1e- = Fe2+

E° = 0.767 V

Cu2+ + 1e- = Cu+

E° = 0.161 V

Part i

At 15 mL of Fe3+ added

VCuMCu > VFeMFe

50*0.05 > 15*0.1

2.5 > 1.5

[Cu2+] = (VFeMFe) / (VCu + VFe)

= (15 mL x 0.1 mol/L x 1L/1000 mL) / [(50+15)mL x 1L/1000 mL)]

= 0.0015 mol / 0.065L

= 0.02308 M

[Cu+] = (VCuMCu - VFeMFe) / (VCu + VFe)

= (50*0.05 - 15*0.1)M*mL/(50+15)mL

= 0.0154 M

Half reaction

Cu2+ + 1e- = Cu+

E° = 0.161 V

From the Nernst equation

E = E° - (0.0592/n) log ([Cu+] / [Cu2+])

= 0.161 - (0.0592/1) log (0.0154/0.02308)

= 0.171 V

Part ii

At 25 mL of Fe3+ added

VCuMCu = VFeMFe

50*0.05 = 25*0.1

2.5 > 2.5

E = (nCuECu + nFeEFe ) / (nCu + nFe )

= (1*0.161 + 1*0.767) / (1+1)

= 0.464 V

Part iii

At 26 mL of Fe3+ added

VCuMCu < VFeMFe

50*0.05 = 26*0.1

2.5 < 2.6

[Fe2+] = VCuMCu / (VCu + VFe)

= (50*0.05)M*mL / (50+26)mL

= 0.0329 M

[Fe3+] = (VFeMFe - VCuMCu ) / (VCu + VFe)

= (26*0.1 - 50*0.05)M*mL/(50+26)mL

= 0.00132 M

From the Nernst equation

E = E° - (0.0592/n) log ([Fe2+] / [Fe3+])

= 0.767 - (0.0592/1) log (0.0329/0.00132)

= 0.684 V