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The correct representation of Nernst`s equaion is .A. `E_(M^(n+) //M) = E_( M^(n+)//M)^@ + n/( 0.591) log (M^(n+))`B. `E_(M^(n+)) //M) = E_(M^(n+)//M)^@ + n/( 0.591) log (M^(n+))`C. `E_(M^(n+) //M) = E_(M^(n+)//M)^@ + n/( 0.591) log (M^(n+))`D. None of the above |
Answer» Correct Answer - C ` E=E^@ - (RT)/(nF) In 1/([M^(n+)]) , E=E^@ + (RT)/(nF) In [M^(n+)]` `E=E^@ + ( 2.303 RT)/(nF) log [M^(n+)]` Substituting the value of `R,T (298 K)` and `F` we get `E=E^@ + ( 0.591)/n log (M^(n+))`. |
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