Calculate the number of coulombs required to deposit 40.5 g of Al when the electrode reaction is , `Al^(3+)+3e^(-)rarr Al`
Calculate the number of coulombs required to deposit 40.5 g of Al when the electrode reaction is ,
`Al^(3+)+3e^(-)rarr Al`
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The charge Q on n moles of electrons is given by Q=nF
`{:(Al^(3+)+3e^(-)rarrAl),(1 "mol" " "3"mole"" "1 "mol"):}`
Thus, charge on 3 mol of electrons `Q =3 molxx96500 " C mol"^(-1)`=289500 C
Molar mass of Al =27g `mol^(-1)`
`:.` To deposit 27 g of Al, the electric charge required =289500 C
To deposit 40.5 g of Al, the electric charge required `=((289500C))/((27 G))xx(40.5 g)=4.342xx10^(5) C`
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