The voltage of a redox reaction involving a titration 300 mL of 0.3M of Fe(H2O)2

Question

The voltage of a redox reaction involving a titration 300 mL of 0.3M of Fe(H2O)2+ against 0.600 M Ce(H2O)7(OH)3+ was monitored in an electrochemistry experiment using a calomel reference electrode. The standard potentials for calomel reference, Fe3+ and Ce4+ reductions are 0.241, 0.767 and 1.70V, respectively. A. Write a balance redox reaction equation for the titration B. What is the oxidizing and reducing agent? C. Calculate the equivalent point of the redox reaction: D. Calculate the cell voltage at 90.0mL

Explanation / Answer

Ecell = Eind - Eref

Ecell=EFe3+/Fe2+ ESCE = 0.767 V 0.241V= +0.526V   

Electrode reaction is:

Ce4+ (aq) + Fe2+ (aq) Ce3+ (aq) + Fe3+ (aq)

Ce+4 or Ce(H2O)7(OH)3+ is the oxidising agent as it helps the oxidation of Fe+2 to Fe+3

and Fe+2 or Fe(H2O)2+ is the reducing agent

Before the equivalence point

Ecell = E0Fe+3/Fe+2 - 0.0591/1 log [Fe+2]/[Fe+3] - Eref

Ecell = 0.767 V 0.241V - 0.0591 log [Fe+2]/[Fe+3]

Ecell = 0.526 - 0.0591 log [Fe+2]/[Fe+3]

At the equivalence point Exactly enough Ce4+ has been added to react with all the Fe2+. This leaves all cerium in the form Ce+3, and all iron is in the form Fe3+.

at the equivalence point [Ce+3] = [Fe+3] and [Ce+4] = [Fe+2]

Now, let us use the electrode potential of both the electrodes

EFe+2/Fe+3= 0.767 - 0.0591 log [Fe+2]/[Fe+3]

ECe+4/Ce+3=1.70 - 0.0591 log [Ce+4]/[Ce+3]

Add these two potentials, 2E = 0.767+1.70 - 0.059 (log [Fe+2]/[Fe+3 ]+ log [Ce+4]/[Ce+3]

= 2.467 - 0.059( log [Ce+3][Fe+2/[Ce+4][Fe+3])

As stated before, [Ce3+] = [Fe3+] and [Ce4+] = [Fe2+] at the eq.point. Hence, the ratio of concentrations in the log term is unity. Therefore, the logarithm is 0 .

2E = 2.467 Or E = 2.467/2 = 1.233 V

Therefore,the cell voltage is 1.233 - Eref  = 1.233 - 0.241 = 0.992 V

equivalence point is when the cell potenial is 0.992 V

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