25 cm^3 of 0.1 mol dm^-3 Cr2Ox^2- was mixed with h2so4 and reacted with excess KI to form iodine and Cr^3+ ions. the iodine liberated reqd 25 cm^3 of 0.6 mol dm^-3 of sodium thiosulfate for reaction. determine the change in oxidation state of Cr and show that x is 7
Originally posted by ohnoez!:25 cm^3 of 0.1 mol dm^-3 Cr2Ox^2- was mixed with h2so4 and reacted with excess KI to form iodine and Cr^3+ ions. the iodine liberated reqd 25 cm^3 of 0.6 mol dm^-3 of sodium thiosulfate for reaction. determine the change in oxidation state of Cr and show that x is 7
Since 0.015 moles of thiosulfate were used, hence 0.015 moles of electrons were transferred. It follows that 7.5x10^-3 moles of iodine were present in the thiosulfate/iodine redox reaction.
Thus 0.015 moles of iodide ions (from KI) was involved in the dichromate/iodide redox reaction. It follows that 0.015 moles of iodide ions donated 0.015 moles of electrons.
Number of moles of dichromate ions present = 2.5x10^-3
Hence number of moles of Cr3+ ions present = 5x10^-3
Since 2.5x10^-3 moles of dichromate ions accepted 0.015 moles of electrons to form 5x10^-3 moles of Cr3+ ions, it follows that :
1 mole of dichromate ions accepted 6 moles of electrons to form 2 moles of Cr3+ ions.
Since each dichromate ion has two chromium atoms, it follows that each mole of chromium atoms must have accepted 3 moles of electrons to form 1 mole of Cr3+ ions.
Therefore the O.S. of chromium must have originally been +6 (the O.S. of Cr in Cr2Ox 2-), and reduced by 3 (since 3 moles of electrons accepted) to +3 (the O.S. of Cr in Cr3+).
Since oxygen is more electronegative than chromium hence the O.S. of oxygen would have been -2, and because the O.S. of chromium we've determined to be +6, hence the formula of the dichromate(VI) ion must have been Cr2O7 2- (ie. the value of x is 7).
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