(a) Calculate the emf for the given cell at `25^@C`:
`Cr|Cr^(3+) (0.1M)||Fe^(2+) (0.01M) |Fe`
(b) calculate the strength of the current required to deposit 1.2g of magnesium from molten `MgCl_2` in 1 hour.

Explanation

`E.M.E=E_(cell)^@-0.059/n log""(Products)/(Reactants)`
`Cr|Cr^(3+) (0.1M)||Fe^(2+) (0.01M)|Fe`
Equation,
`2Cr+3Fe^(2+)to 2Cr^(3+)+3Fe`
`E_(Cr^(3+)//Cr)^@=-0.74E_(Fe^(2+)//Fe)^@=-0.44V`
`E_(cell)^@=E_(cathode)-E_(anode)`
`=-0.44-(-0.74)`
=0.3
`E_(cell)=0.3-0.059/n log""(0.1)^2/(0.01)^3`
`=0.3-0.0098 log""0.01/0.00001`
`=0.3-0.0098 log 10^4`
`=0.3-0.0393`
`=+0.260 V`
(b) The half reaction occuring during deposition is:
`Mg^(2+)+2e^(-) to Mg(s)`
So to produce 1 mole of Mg we need 2 moles of electrons,
1 mol of Mg=24 g
Thus 1.2g of Mg=(`1.2//24`)=0.05 moles
Hence no. of moles of electrons required to deposit
`=0.05 times 2=0.1 mol es`
Hence no. of moles of electrons required to deposit
`=0.05 times 2=0.1 mol es`
Qty. of electricity passed (Q)= moles of electrons
`passed times 96500`
`=0.1 times 96500=9650 C`
`I=Q//t`
`I=9650/(60 times 60(converti ng 1 hr into seconds))`
`I=9650/3600=2.68A`