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)` ?

Options

A. `3`
B. `6`
C. `2`
D. `7`

Answer

A

Explanation

`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`.