Question

Sulphur dioxide gas and acidified potassium dichromate equation

Answer 1

Sulphur dioxide gas (SO2) is bubbled through acidified potassium dichromate (K2Cr2O7/H+). The orange solution turns green.

Half equations: Cr2O72- (aq) +14H+ (aq) + 6e- = 2Cr3+ (aq) + 7H2O (l) and SO2 (g) + 2H2O (l) = SO42- (aq) + 4H+ (aq) +2e-

Full equations: Cr2O72- (aq) + 2H+ (aq) + 3SO2 (g) = 2Cr3+ (aq) + H2O (l) + 3SO42- (aq)

Colour changes: K+ in K2Cr207 is a colourless ion, and dichromate (Cr2O72-) is an orange solution. SO2 is a colourless gas, so the intial solution is orange.
The solution turns green because the Cr2O72- is reduced to Cr3+ which is a green ion- so the solution turns from orange to green.

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Answer 2

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