Q. The charge required for reduction of 1 mole of ${Cr}_{2}O_{7}^{2 -}$ ions to ${Cr}^{3 +}$ is
A) 96500 C
B) 2 × 96500 C
C) 3 × 96500 C
D) 6 × 96500 C
Explanation
(b)
${Cr}_{2}O_{7}^{2 -} + \ {6e}^{-} \longrightarrow {2Cr}^{3 +}$
Reduction of 1 mol of ${Cr}_{2}O_{7}^{2 -}$ to ${Cr}^{3 +}$ required 6 moles of electrons. Hence, charge required = 2 × 96500 C
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.