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The half cell potentials of a half cell A^((x+n)^(+)),A^(x+)|Pt were found to be as follows {:("% of reduced form",24.4,48.8),("Half cell potential (V)",0.101,0.155):} The value of n is:-

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Solution :Considering the REDUCTION reaction
`UNDERSET("Oxidised form")(A^(X+n)+)+n eto underset("Reduction form")(A^(x+))`
`therefore[A^(x+)]=24.4""therefore[A^((z+n)+)]=75.6,""E_(RP)=0.101V`
Now from NERNST's equation
`E_(RP)=E_(RP)^(o)+(0.059)/(n)"log"([OF])/([RF])`
`0.101=E_(RP)^(o)+(0.059)/(n)"log"(75.6)/(24.4)` . . . (i)
if `[A^(x+)]=48.8[A^((x+n)+)]=51.2,E=0.115V`
`therefore0.115=E_(RP)^(o)+(0.059)/(n)"log"(51.2)/(48.8)` . . . (i8i)
By eqs. (II)-(i),
`therefore0.014=(0.059)/(n)["log"(51.2)/(48.8)-"log"(75.6)/(24.4)]`
`thereforen=(0.059)/(0.014)[0.49-0.020]`
`n=1.98=2`.


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