When a solution of potassium iodide is added to acidified potassium dichromate, a colour change of orange to green is observed.
Write the redox half-equations and hence the balanced redox equation for the reaction taking place.

It's a redox equation in which potassium iodide(KI) is being oxidized to Iodine (I2) while potassium dichromate is reduced to Chromium (III) (Cr3+) and such we have to first break them into two half reactions. One for the substance being oxidized and the other for that which is being reduced.

Explanation:

Going straight to the half reactions:

2KI = 2K(+) + I2 + 2e-

and

K2Cr2O7 + 14H(+) + 6e- = 2K(+) + 2Cr3(+) + 7H20

Inspecting the two equations above, we see that the electrons produced by KI during oxidation is 2 while that produced by K2Cr2O7 is 6.

We have to make them equal. Therefore, we multiply each term in the oxidation equation by 3.

We have:

6KI = 6K(+) + 3 I2 + 6e-

For the reduction equation, the 14H(+) has to be broken down due to the fact that this was mixed in a sulphuric acid (H2SO4).

With that in mind, rebalancing the reduction equation, we have:

K2Cr2O7 + 7H2SO4 + 6e- = 2K(+) + 2Cr3(+) + 7H20 + 7SO4(2-)

Now, we add the new oxidation and reduction equations together to get:

6KI + K2Cr2O7 + 7H2SO4 + 6e- = 8K(+) + 3I2 + 6e(-) + 2Cr3(+) + 7H20 + 7SO4(2-)

Simplifying it, the 6e- cancels each other to give us:

6KI + K2Cr2O7 + 7H2SO4 = 8K(+) + 3 I2 + 2Cr3(+) + 7H20 + 7SO4(2-)

Now Simplifying further, the SO4(2-) is going to act as a counter ion for the K(+) and Cr(3+).

Therefore,we now have the balanced redox equation as:

6KI + K2Cr2O7 + 7H2SO4 = 4K2SO4 + 3 I2 + Cr2SO4(3) + 7H20

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