Potassium iodide reacts with acidified `K_(2)Cr_(2)O_(7)`. How many moles of KI are required for one mole of `K_(2)Cr_(2)O_(7)` ?A. `3`B. `6`C. `2`D. `7`

Potassium iodide reacts with acidified `K_(2)Cr_(2)O_(7)`. How many mo

1.	Potassium iodide reacts with acidified `K_(2)Cr_(2)O_(7)`. How many moles of KI are required for one mole of `K_(2)Cr_(2)O_(7)` ?A. `3`B. `6`C. `2`D. `7`
Answer» Correct Answer - A `overset(+6)(C)r_(2)O_(7)^(2-)+6e^(-) rarr 2Cr^(3+)` According to the reduction half-reaction, `1` mol of ` K_Cr_2O_7` gains ` 6` mol of electrons. ` 2I^(-) rarr I_2 = 2e^(-)` To produce `1` mol of `I_2` , 2 mol of electrons are lost. Thus ,by gaining ` 6` mol of electrons , 1 mol of ` K-2Cr_2 O_7` will liberate `3` mol of ` I_2` ` Cr_2O_7^(2-) + 14H^(+) = 6I^(-) rarr 2Cr^(3+) + 7H_2O + 3I_2`.

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Potassium iodide reacts with acidified `K_(2)Cr_(2)O_(7)`. How many moles of KI are required for one mole of `K_(2)Cr_(2)O_(7)` ?A. `3`B. `6`C. `2`D. `7`

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