The EMF of the cell
`Mg| Mg^(2+) (0.01M) | | Sn^(2+) (0.1M) | Sn` at 298 K is
(Given `E_(Mg^(2+) | Mg)^(@) = -2.34 V , E_(Sn^(2+) | Sn)^(@) = -0.14` V)

Options

A. `2.17` V
B. `2.23` V
C. `2.51` V
D. `2.45` V

Answer

B

Explanation

The redox reaction is
`{:("Mg" rarr "Mg"^(2+)+2e^(-)("oxidation")),("Sn"^(2+)+2e^(-) rarr "Sn"("Reduction")),(Mg+"Sn"^(2+) rarr "Mg"^(2+) + "Sn"):}`
`E_("cell") = E_(Sn)^(@) - E_(Mg)^(@)`
`= -0.14 - (-2.34) = 2.2` V
According to Nernst equation
`E_(cell) = E_("cell")^(@) - (0.0591)/(n) "log"_(10) ([P])/([R])`
`= E_("cell")^(@) - (0.0591)/(n) "log" ([Mg^(2+)])/([Sn^(2+)])`
`(because [Sn_((s)) ] = [Mg_((s))] = 1)`
`therefore E_("cell") (2.2V) - ((0.0591V)/(2) "log" (1)/(10))`
`= (2.2V) - ((0.0591)/(2) V) log 10 = 2.23` V