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Electrolysis of Aqueous Potassium Iodide
Transcript of Electrolysis of Aqueous Potassium Iodide
a 1 mol/L aqueous solution of potassium iodide?
Are the observed products the ones predicted
using reduction potentials? Predictions O + 4H + 4e --> 2H O 2 (g) + - 2 (l) E = 0.815 V 0 1 I + 2e --> 2I (s) - - (aq) 2 2 E = 0.536 V 0 Oxidation 2H O + 2e --> H + 2OH (l) 2 (g) - (aq) E = -0.414 V 0 3 K + e --> K + (aq) - (s) E = -2.931 V 0 Reduction 4 3-1: -1.229 V 4-1: -3.746 V 3-2: -0.95 V 4-2: 3.467 V From this, we predict that hydrogen gas, hydroxide ions and iodine are produced. Materials U-tube
600 mL beaker
Sheet of paper
Pipettes Copper wire rod
50 mL of 1 M KI solution
Power source Experiment! What do we notice? Iodine is produced Hydroxide ions are produced + - > > Anode Cathode K --> + <-- I - 2I --> I + 2e - (aq) 2 (s) - 2 - 2H O + 2e --> H + 2OH 2 (l) - 2(g) - (aq) ANODE CATHODE Overall: 2H O + 2I --> I + H + 2OH 2 (l) 2(g) - (aq) - (aq) 2(s) 0.95 volts were predicted.
Much more was required because of the concept of overvoltage. Conclusion? The products were indeed hydrogen gas, hydroxide ions, and solid iodine (as we predicted) If you repeated the electrolysis using aqueous
sodium iodide instead of aqueous potassium
iodide, would your observations change? No they would not. The reduction potential of sodium ions is similar to potassium ions (-2.711 to -2.931), and the same products can be predicted. To make potassium by electrolyzing potassium iodide, would you need to modify the procedure? The only alternative that we have discussed is using molten potassium iodide. To produce this, you must keep potassium iodide at a very high temperature.