Calculate the value of `E_("cell")` at 298 K for the following cell:
`Al//Al^(3+) (0.01M)"||" Sn^(2+) (0.015 M)//Sn`
`[E_(Al^(3+)//Al)^(Theta) = -1.66V and E_(Sn^(2+)//Sn)^(Theta) = -0.14V]`

Explanation

Given cell is :
`Al/ /A1^(3+) (0*01 M) ।। Sn^(2+) (0*015 M)//Sn`
given, `E_(AI^(3+)//AI)^@ = – 1*66V`,
`E_(Sn^(2+)//Sn)^@ = -0*14V`
`:. " " E_(cell)^@ = E_(R)^@ - E_(L)^@`
`= -0*14 - (-1*66)`
`= -0*14+ 1*66`
`= 1*52V`
and net cell reaction is
`2Al + 3Sn^(2+) to 2AI^(3+) + 3Sn`
so net cell reaction involves transfer of 6 mole of electrons
`:. " " n = 6`
According to Nernst equation, at 298K
`E_(cell) = E_(cell)^@ -(0*059)/n log "([AI^(3+)]^2)/([Sn^(2+)]^3)`
`=1*52 -(0*059)/6 log " ([0*01]^2)/([0*015]^3)`
`=1*52 - 0*01447`
`E_(cell) = 1*50553V`.