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]`

Explanation

The half-cell reaction is-
`A^(x+) rarr A^((x+n)x) +"ne"^(-)`
Its Nernst equation is-
`E = E^(@) -(RT)/(nF)log.([A^((xn)+)])/([A^(x+)]) = E^(@) -((0.06V)/(n)) log.(("oxidized form")/("reduced form"))`
Substituting the given values, we get
`0.101 V = E^(@) - ((0.06V)/(n)) log.(75.6)/(24.4) = E^(@) -((0.06V)/(n)) (0.491)` ..(i)
`0.115V = E^(@) -((0.06V)/(n)) log.(51.2)/(48.8) =E^(@) -((0.06V)/(n))(0.021)` ..(ii)
eq.(ii) -(i)
`0.014 = ((0.06V)/(n)) (0.491 -0.021)`
or `n = ((0.06V)(0.47))/((0.014V)) ~=2`