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Write the Nernst equation and emf of the...

Write the Nernst equation and emf of the following cell at 298 K.
(i) `Mg(s) |Mg^(2+) ( 0.001 M) || Cu^(2+) ( 0.0001M ) | Cu(s)`
(ii) `Fe(s) | Fe^(2+) ( 0.001M) || H^(+) ( 1M) | H_(2)(g) (1 "bar") | Pt(s)`
Given that, `E_(Mg^(2+) //Mg)^(@) = - 2.36V`,
`E_(Cu^(2+) Cu)^(@) = 0.34V, E_(Fe^(2+)Fe)^(@) = -0.44V`

Text Solution

Verified by Experts

(I) For the given cell, the half-cell reactions will be given as below `:`
At anode `Mg rarr Mg^(2+) + 2e^(-)`
At cathode `Cu^(2+) + 2e^(-) rarr Cu`
Therefore, the overall cell reaction will be
`Mg+ Cu^(2+) rarr Mg^(2+) + Cu`
The Nernst equation is
`E_("cell") = E_("cell")^(@) - ( 0.0591)/( n ) log ""([Mg^(2+)])/([Cu^(2+)])`
In this case, n =2
Therefore,
`E_("cell") = [ 0.34 - ( - 2.36)] - ( 0.0591)/( 2) log""((0.001))/((0.0001))`
`= 2.7 - 0.0295 log ( 10) = 2.7 - 0.0295`
`= 2.67 V` ( approx.)
(ii) `Fe(s) | Fe^(2+) ( 0.001M) || H^(+) ( 1M ) | H_(2)( g) ( 1" bar" | Pt (s)`
The cell reaction is `Fe+ 2H^(+) rarr Fe^(2+) + H_(2)`
The Nernst equation is
`E_("cell") = E_("cell")^(@) - ( 0.059)/( n ) log ""([Fe^(2+)])/([H^(+)]^(2))`
In this case, n =2
Therefore, `E_("cell") = [ 0- ( - 0.44) ] - ( 0.0591)/( 2) log ""(0.001)/( (1)^(2))`
`= 0.44 - ( 0.0295 ) xx log ( 10^(-3))`
`= 0.44 - 0.0295 xx ( -3)= 0.44 + 0.0885`
`= 0.5285 V = 0.85V` ( approx.)
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