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+)` ?

How much charge is required to reduce (a) 1 mole of `Al^(3+)` to Al an

1.	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+)` ?
Answer» (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

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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+)` ?

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