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# Complete the following chemical equations :(a) KI + CI2 →      (b) KcIO3­ + I2 →Justify the formation of the products in the above reactions.

Navjyot Kalra
6 years ago
(a) 2KI + CI2 → 2KCI + I2
Since CI2 is more powerful oxidizing agent than I2, CI­2, is able to displace I- to form I2.
2I- → I2 + 2e-, E0 = + 0.54 V ….(i)
CI2 + 2e- → 2CI- E0 = 1.36V ….(ii)
On subtracting eq. (i) from eq. (ii), we get
$CI_{2_{g}} + 2I_{(aq)}^{-} +\rightarrow 2CI_{(aq)}^{-} + I_{2_{(s)}} E^{0} = 82V$
(b) 2KCIO3 + I2 → 2KIO3 + CI2
Here $CIO_{3}^{-}$is more powerful oxidizing agent than $IO_{3}^{-}$, so CI is displaced by I.
$2IO_{3}^{-}$+ 12H+ + 10e- → I2 + 6H2O, E0 = 1.195V ….(i)
$2CIO_{3}^{-}$+ 12H+ + 10e- → CI2 + 6H2O, E0 = 1.47V ….(ii)
On substracting eq. (i) from eq. (ii), we get
$2CIO_{3}^{-}$+ I2 $2IO_{3}^{-}$+ CI2 E0 = 0.275V