Question

How many Faradaye of charge are required to reduce \( 1 mol \) of \( Cr _{2} O _{7}^{2} \) to \( Cr ^{+3} \) ? [NCERT \( Pag \), B8] (1) \( 6 F \) (2) \( 3 F \) (3) \( 2 F \) (4) \( 1 F \)

Answer 1

Answer: 3 F

Cr has a formal charge of +6 as 

[(Cr⁺⁶)₂(O^-2)₇]^-2

So in this half cell reaction,

Cr⁺⁶  + 3e^-→  Cr⁺³

There are 3 electrons participating in this reaction. 

Also note that, 1 Farad is the charge when 1 mole of electrons are taking part in the reaction, so when 3 moles of electrons take part, 3 Farad of charge is required.

(3 electrons in reaction means 3 moles of electrons when expressed in moles).