The half cell potential of a half-cell `A^(x+), A^((x+n)+)|Pt` were found to be as follows:
`{:(%"of reduced form",24.4,48.8),("Cell potential"//N,0.101,0.115):}`
Determine the value of `n`.
[take `(2.303RT)/(F) = 0.05, log_(10)24.4 = 1.387, log_(10) 75.6 = 1.878 log_(10) 48.8 = 1.688, log_(10) 51.2 = 1.709]`

Answer

2

Explanation

Considering the reduction reaction
`A^((x+n)+)+ "ne" to A^(x+)`
`:. overset("Oxidized form")([A^(x+)]=24.4) :. overset("Reduced form")([A^((x+n)+)])=75.6, E_(RP)=0.10V`
Now from Nernst's equation
`E_(RP)=E_(RP)^(0)+0.059/n "log" ([OF])/([RF])`
`0.101=E_(RP)^(0)+0.059/n "log" 75.6/24.4`...........(i)
If `]A^(x+)]=48.8, [A^((x+n)+)]=51.2, E=0.115`
`:. 0.005=E_(RP)^(0)+0.059/n "log" 51.2/48.8`..........(ii)
By eqs. (ii) -(i)
`:. 0.0014=0.059/n["log" 51.2/48.8-"log"75.6/24.4]`
`=n=0.059/0.014[0.49-0.020]`
`n=1.98~~2`