a) BaI2(aq) b) NiSO4(aq) c) LiF(aq) Solution (a) In the molten form of barium io

Question

a) BaI2(aq)
b) NiSO4(aq)
c) LiF(aq)

Explanation / Answer

(a) In the molten form of barium iodide, barium and iodine ions are present: BaI2(l) Ba^2+(l) + 2I^-(l) When electricity is applied, Ba^2+ ions move to the negative electrode (cathode) where they are reduced to metallic barium by gaining 2e-; Ba^2+(l) + 2e- --------> Ba(s) I^- ions move to the positive electrode (anode) where they are oxidized to elemental iodine by losing 2e-; 2I^-(l) -------> I2(s) + 2e- (b) At cathode: Ni^2 +2e- = Ni At Anode : OH^-1 -1e- = OH x4 4OH=2H2o + O2 Nickel ion being of higher concentration gets reduced more readily than hydrogen. At anode, the OH ion being lower in the electrochemical series gets oxidized to liberate oxygen. If it is less concentrated : At cathode: H^ +1e- =H x4 2H+ 2H= 2H2 At anode: OH^- -1e- = OH x4 4OH= 2H2o +O2 At cathode as hydrogen is a simple ion compared to nickel it gets discharged first At anode the same as above (c) The electrolysis of LiF will usually lead to the production of Li metal and molecular flourine 2LiF -> 2Li + F2 where the half cell reactions, with reduction potentials, are Li+ + 1e- -> Li, Eored = -3.05 V 2F- -> F2 + 2e-, Eored = +2.87 V However, the presence of water will provide competing reactions 2H20 + 2e- -> H2 + 2OH-, Eored = -0.83 V 2H20 -> O2 + 4H+ + 4e-, Eored = +1.23 V

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