How much charge is required to reduce (a) 1 mole of `Al^(3+)` to Al and (b) 1 mole of `MnO_(4)^(-)` to `Mn^(2+)` ?

Explanation

(a) The reduction reaction is :
`underset("1 mole")(Al^(3+))+ underset("3 mole")(3e^(-)) rarr Al`
Thus, 3 mole of electrons are needed to reduce 1 mole of `Al^(3+)` `Q=3xxF`
`=3xx96500=289500` coulomb
(b) The reduction reaction is :
`underset("1 mole")(MnO_(4)^(-))+8H^(+) + underset("5 mole")(5e^(-)) rarr Mn^(2+)+ 4H_(2)O`
`Q=5xxF`
`=5xx96500=482500` coulomb