1.

Given E_(Cr^(3+)//Cr)^(o)=-0.72V,E_(Fe^(2+)//Fe)^(o)=-0.42V. The potential for the cell Cr|Cr^(3+)(0.1M)||Fe^(3+)(0.01M)|Fe is

Answer»

0.339V
`-0.339V`
`-0.26V`
`0.26V`

Solution :`E_(CELL)=E_(cell)^(o)-(0.059)/(n)log" "Q`
`=(-0.42+0.72)-(0.059)/(6)"log"((0.1)^(2))/((0.01)^(3))([CrtoCr^(3+)+3overline(e)]xx2,[Fe^(2+)+2overline(e)toFe]xx3)/(2Cr+3Fe^(++)to2Cr^(3+)+3Fe`
`E_(cell)=0.30-(0.059)/(6)"log"(10^(-2))/(10^(-6))=0.30-(4xx0.059)/(6)LOG10`
`=0.30-0.0393=0.2607`VOLT.


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