Q. One mole of acidified $K_{2}{Cr}_{2}O_{7}$ on reaction with excess KI will liberate………mole(s) of $I_{2}.$
A) 6
B) 1
C) 7
D) 3
Explanation
(d)
${Cr}_{2}O_{7}^{2 -} + 14H^{+} + 6e^{-} \longrightarrow 2{Cr}^{3 +} + 7H_{2}O$
$\left( 2I^{-} \longrightarrow I_{2} + 2e^{-} \right) \times 3$
[table] ${Cr}_{2}O_{7}^{2 -} + 14H^{+} + 6I^{-} \longrightarrow 2{Cr}^{3 +} + 7H_{2}O + 3I_{2}$ [/table]
Hence, number of moles of $I_{2}$produced =3
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.