Question

Titrate 100.0mL of 0.0500M Fe2+ with 0.100M Ce4+. The equivalence point is when volume of Ce4+ added is 50.0mL. Find the voltage of the cell at the following points: a. 20.0mL b. 36.0mL c. 50.0mL d. 51.0mL e. 63.0mL

Answer 1

Titrate 100.0mL of 0.0500M Fe2+ with 0.100M Ce4+. The equivalence point is when volume of Ce4+ added is 50.0mL. Find the voltage of the cell at the following points: a. 20.0mL b. 36.0mL c. 50.0mL d. 51.0mL e. 63.0Ml

a. 20.0mL

At 20.0 ml

This is 20.0/50.0 of the point to the equivalence point, hence 20.0/50.0 of the iron is present as Fe3+ and 30.0/20.0 is in the form of Fe2+

So [Fe2+] /[Fe3+]= 30.0/20.0 = 30.0/20.0

                             20.0/50.0

E = 0.526 -0.0591 log {[Fe2+] /[Fe3+]}

= 0.526 -0.0591 log20.0/50.0

= 0.526 -0.0591 *(-0.398)

= 0.550 V

At 36.00 ml

This is 36.0/50.0 of the point to the equivalence point, hence 36.0/50.0 of the iron is present as Fe3+ and 14.0/36.0   is in the form of Fe2+

So [Fe2+] /[Fe3+]= = 14.0/36.0

                            

E = 0.526 -0.0591 log {[Fe2+] /[Fe3+]}

= 0.526 -0.0591 log14.0/36.0

= 0.526 -0.0591 *(-0.410)

= 0.550 V

At 50.0 ml

This is the equivalent point and the cell voltage at this point is 0.99 V

Which is calculating as follows:

E = E+ – E (calomel)

= 1.23-0.241

= 0.99 V

Noe after equilvalent point

After 51..0 ml

Here Cerium were converted into Ce2+. There is excess of 1.0 ml of Ce4+ so [Ce2+] / [Ce4+] = 50.0 /1.00

Therefore;

E= E+- E calomal

= 1.70 – 0.0591 log [Ce3+] / [Ce4+]

= 1.70 – 0.0591 log 50.0 /1.00

= 1.70- 0.0591* (1.69)

= 1.60 V

After 63.0 ml

Here Cerium were converted into Ce2+. There is excess of 13 ml of Ce4+ so [Ce2+] / [Ce4+] = 50.0 /13.00

Therefore;

E= E+- E calomal

= 1.70 – 0.0591 log [Ce3+] / [Ce4+]

= 1.70 – 0.0591 log 50.0 /13.00

= 1.70- 0.0591* (0.585)

= 1.66 V