A mixture of starch solution and potassium iodide was placed in a test tube. On adding dilute tetraoxosulphate (lV) acid and then K2Cr2O7 solutions, a blue-black colour was produced. In this reaction, the?
iodide is oxidized
tetraoxosulphate (Vl) acid acts as an oxidizing agent
starch has been oxidized
K2Cr2O7 is oxidized
Explanation
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Discussions (8)
Explanation:
In this reaction, a mixture of starch solution and potassium iodide is first prepared. Then, dilute tetraoxosulphate (IV) acid (H2SO4) and potassium dichromate (K2Cr2O7) solutions are added. The reaction produces a blue-black color, which indicates the formation of a starch-iodine complex.
The blue-black color forms because the iodide ions (I-) are oxidized to iodine (I2) in the presence of an oxidizing agent. In this case, the oxidizing agent is the potassium dichromate (K2Cr2O7). The iodine then reacts with the starch molecules, forming the blue-black starch-iodine complex.
So, in this reaction, the iodide is oxidized (Option A). However, the correct answer is marked as Option C, which states that starch has been oxidized. This is not accurate, as starch is not undergoing oxidation in this reaction. Instead, it is reacting with the iodine produced by the oxidation of iodide ions. The correct answer should be Option A, not Option C.
the answer is a coz it's the iodide that was oxidized to iodine which turned the starch blue black
The correct answer is A: iodide is oxidized.
Explanation:
The reaction involves potassium dichromate (K2Cr2O7) oxidizing iodide ions (I-) to form iodine (I2), which then reacts with starch to produce the blue-black color.
Step-by-step:
1. Iodide ions (I-) are oxidized to iodine (I2) by K2Cr2O7.
2. Iodine reacts with starch to form the blue-black complex.
Conclusion:
In this reaction, iodide is indeed oxidized, resulting in the color change.
