Oxidation number of Cr in K₂Cr₂O₇ & K₂Cr₂O₄ is:
Correct answer: D. +6, +4
- A. +6, +3
- B. +5, +2
- C. +7, +3
- D. +6, +4
Explanation
K₂Cr₂O₇ (Potassium Dichromate):The oxidation number of potassium (K) is always +1.The oxidation number of oxygen (O) is usually -2.Let the oxidation number of chromium (Cr) be x.The overall charge of the compound is 0.Therefore, 2(+1) + 2(x) + 7(-2) = 0Solving for x, we get x = +6.K₂Cr₂O₄ (Potassium Chromate):Using the same logic as above, we can calculate the oxidation number of chromium in potassium chromate:2(+1) + 2(x) + 4(-2) = 0Solving for x, we get x = +4.
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Electrochemistry describes electron transfer through oxidation and reduction, identifies oxidizing and reducing agents, and applies oxidation numbers to balance redox equations. It also covers electrode potential and the standard hydrogen electrode, which provides the reference for comparing the reduction tendencies of other electrodes.
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