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Number of electron involved in the reduc...

Number of electron involved in the reduction of `Cr_(2)O_(7)^(2-)` ion in acidic solution to `Cr^(3+)` is:

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The correct Answer is:
D
`underset("reduction")(Cr_(2)O_(7)^(2-) + 6e^(-) to 2Cr^(3+)) implies 6e^(-)` are required for 1 mol of `Cr_(2)O_(7)^(2-)`
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VMC MODULES ENGLISH-STOICHIOMETRY-II-LEVEL (1)
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