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Calculate the potential ofa Daniel cell, initially containing 1 litre each of 1M Cu^(2+) and 1M Zn^(2+) after a passage of 1 times 10^5 coulombs of charge . E_(Cu^(2+),Cu)^@=+0.34V, E_(Zn^(2+),Zn)^@=-0.76V |
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Answer» Solution :Moles of ELECTRICITY PASSED `=10^5/96500=1.04F` `therefore` eq. of `Cu^(2+)` removed=1.04 or mole of `Cu^(2+)` removed=0.52 and mole of `Zn^(2+)` PRODUCED= 0.52 Thus, `[Zn^(2+)]`=1.52 M and `[Cu^(2+)]=0.48M` For the Daniel cell, `Cu^(2+)+Zn=Cu + Zn^(2+)` `E=1.10 -0.0591/2 LOG"" 10.52/0.48=1.09V` |
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