- Explain briefly what was occurring in Table 2 and write the balanced net ionic equations
for reactions which occurred.
“Like dissolves like” is a fundamental rule of solubility. Mineral oil is nonpolar. Therefore,
one would expect nonpolar solutes to dissolve in it. Halide ions are charged particles,
therefore one would not expect them to dissolve in a nonpolar solvent like mineral oil, but
should dissolve in water, a polar solvent. From the data, it appears that the nonpolar halo-
gens dissolved in the nonpolar mineral oil. When aqueous solutions of halide ions and
halogens are mixed together, the color of the mineral oil layer should indicate whether a
reaction had occurred or not. If the color of the halogen appeared in the mineral oil from
what initially began as the halide ion, then one can conclude that the halide ion was oxi-
dized to the halogen; i.e. 22 Br()--aq()colorless "Br 2 (),+ e (orangish-brown). The mixtures
in which reactions occurred were:
Br 22 ()aq++ 22 I--()aq" Br ()aq I ()aq
Cl 22 ()aq++ 22 Br--()aq" Cl ()aq Br ()aq
Cl 22 ()aq++ 22 I--()aq" Cl ()aq I ()aq
The most reactive halogen was chlorine, followed by bromine and then iodine which ap-
peared to not react. This order agrees with their order of reactivity in the periodic table.- In reference to Table 4 above and using a chart of E^0 redpotentials, determine the potentials
had hydrogen been used as a reference electrode instead of the Zn electrode and calculate
the differences.
Reduction Rxn Voltages E^0 redVoltages Difference
Using Zn Using H 2 (Chart)
Ag+-+e "Ag 1.40 0.80 0.60
Cu^2 +-+ 2 e "Cu 0.99 0.34 0.65
Fe^3 ++ 3 e-"Fe 0.55 −0.04 0.59
Mg^2 +-+ 2 e "Mg 0.60 −2.37 2.97
Pb^2 +-+ 2 e "Pb 0.48 −0.13 0.61
Zn^2 +-+ 2 e "Zn 0.00 −0.76 0.76
- For each reaction studied in Part II, Table 5, write a balanced redox equation.
()aq+ ()aq
Cu ++Fe Cu Fe
()ss()(^22) " +
()aq+ ()aq
Pb++Mg Pb Mg
()ss()
(^22) " +
()aq+ ()aq
Cu ++Pb Cu Pb
()ss()
(^22) " +
Laboratory Experiments