InterviewSolution
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InterviewSolution
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(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. |
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Answer» Solution :`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` |
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