Question

Determine the E in each titrant volumes: 10.00 mL; 20.00 mL; and 30.00 mL, for the titration
of 20.0 mL of 0.100 M Fe^2+ with 0.100 M Ce^4+ in a matrix of 1 M HClO4.

In 1 M HClO4:
E^o (Fe^3+/Fe^2+) = +0.767 V
E^o (Ce^4+/Ce^3+) = +1.70 V

Answer 1

$Fe^{2+} + Ce^{4+} \to Fe^{3+} + Ce^{3+}$

oxidation half reaction

$Fe^{2+} \to Fe^{3+} + e^-$ $E^o= -0.767 V$

reduction half reaction

$Ce^{4+} + e^- \to Ce^{3+}$ $E^o = +1.70 V$

Potential after addition of 10.00 mL of $Ce^{4+}$.

$[Fe^{3+}] =\dfrac{10 × 0.100}{20×0.100} - [Ce^{4+}]=1/2$

$[Fe^{2+}] =\dfrac{(20×0.100)-(10 × 0.100)}{30} + [Ce^{4+}]= 1/30$

Substitution into Nernst equation:

$E_{system} = 0.767 - \dfrac{0.0592}{1}log\dfrac{1/30}{1/2} =$ 0.837V

$E _{eq} = \dfrac{E^f_{Ce^{4+}} + E^f_{Fe^{3+}}}{2} = \dfrac{1.70+ 0.767}{2} = 1.234V$