Answer to Question #150301 in Analytical Chemistry for name123

Question #150301
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
1
Expert's answer
2020-12-29T02:42:39-0500

"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 \u00d7 0.100}{20\u00d70.100} - [Ce^{4+}]=1\/2"


"[Fe^{2+}] =\\dfrac{(20\u00d70.100)-(10 \u00d7 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\r_{eq} = \\dfrac{E^f_{Ce^{4+}} + E^f_{Fe^{3+}}}{2} = \\dfrac{1.70+ 0.767}{2} = 1.234V"


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