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The half cell potentials of a halfcell ...

The half cell potentials of a halfcell `A^((x+n)+) , A^(x+)|pt` were found to be as follows : `{:(% "of reduced form",24.4,48.8),("Half cell potential (V)",0.101,0.115):}`
Determinwe the value of `n`.

Text Solution

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The correct Answer is:
B
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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