1.

3 gm of copper metal in cathode is deposited on electrode when 3 ampere current is pass for 2 hours through aqueous solution of CuSO_(4). Then what is the efficiency of current ? (Atomic weight of Cu=63.5 gm/mol)

Answer»

0.33
`48.7%`
`42.2%`
`54.4%`

Solution :Coulomb=Ampere`xx`Second
`=3xx7200`
`=(3xx7200)/(96500)=(3xx72)/(965)F`. . . (i)
MOLE=`("weight")/("MOLECULAR weight")=(3gm)/(63.5" gm mole"^(-1))`
`Cu_((aq))^(2+)+2e^(-) to Cu_((S))`
So, 2F gives 1 mole Cu . . . (Cathodic reaction)
So, `(3xx72)/(965)` F Gives
Theoretical mole of Cu
`=(3xx72)/(965)F((1" mole")/(2F))=(3xx36)/(965)`
So, efficiency
`=("PRACTICAL value")/("theoretical value")=(3)/(63.5)xx(965)/(3xx36)=0.4221`
`=42.21%`


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