Redox and redox equilibria
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I am doing a titration between iodine and sodium thiosulphate, and I need to explain why the iodine turns colourless when reduced. Why is iodine a brownish colour in solution and why are I- ions colourless? Also how does starch indicate iodine? What makes it turn blue/black?
Corrie writes ....
The 'natural' colour of iodine, as a vapour or in solution in a non-polar solvent, is purple. This means the iodine (I2) molecules are absorbing green light, and letting through red and blue light. The energy of green light corresponds to the energy needed to excite an electron in the iodine molecule to a higher energy state.
In polar solvents, such as water, iodine is reddish-brown, as you say. The interaction of the polar water molecules with the iodine changes the energy levels in the iodine molecules, and hence the colour of light absorbed.
Iodine ions (I-) will have an extra electron and differently spaced set of electron energy levels from iodine molecules. The energy required to excite an electron is in the ultraviolet region. So iodide ions do not absorb visible light and therefore appear colourless in solution.
Starch is a polymeric substance which 'reacts' reversibly with iodine molecules to form an intensely coloured blue-black 'complex', the structure of which is not very well understood. It is thought that somehow the iodine molecules are caught up in the spiral chains formed by the starch molecules. If so-called 'soluble starch' is used, this complex is water soluble and can be used to indicate the presence of very low concentrations of iodine in water.
Once again the colour of the iodine is changed because its electronic energy levels have changed slightly in the 'complex'. As a result, in addition to green, more red light is absorbed, leaving mainly blue light to pass through.
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updated: 05 February 2007
