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Write Nernst equation for the electrode ...

Write Nernst equation for the electrode reaction:
`M^(n+) +n e^(-) to M(s)`

Text Solution

Verified by Experts

`E_(M^(n+)//M)=E_(M^(n+)//M)^(@)+(2.303RT)/(nF)log(([Mn^(n+)])/([M]))`
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Knowledge Check

  • For a cell reaction : M^(n+)(aq)+"ne"^(-) to M(s) , the Nernst equation for electrode potential at any concertation measured with respect to standard hydrogen electrode is represented as

    A
    `E_((M^(n+)//M))=E_((M^(n+)//M))^@-"RT"/"nF"ln (1/([M^(n+)]))`
    B
    `E_((M//M^(n+)))=E_((M//M^(n+)))^@-"RT"/"nF"ln (([M^(n+)])/([M]))`
    C
    `E_((M^(n+)//M))=E_((M^(n+)//M))^@-"RT"/"nF" "log" 1/([M])`
    D
    `E_((M^(n+)//M))=E_((M^(n+)//M^(n+)))^@-"RT"/"nF"ln [M^(n+)]`
  • The correct Nernst equation for the given cell Pt_((s)) | Br_(2(l)) | Br^(-) (M) || H^(+) (M) |H_(2(g)) (1bar) | Pt_((s)) is

    A
    `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([Br_(2(l))][H_2])/([H^+]^2[Br^-]^2)`
    B
    `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([H^+]^2[Br^-]^2)/([Br_(2(l))][H_2])`
    C
    `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([H^+]^2[H_2])/([Br_(2(l))][Br^(-)]^2)`
    D
    `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([Br_(2(l))][Br^-]^2)/([H^+]^2[H_2])`
  • Mark the correct Nernst equation for the given cell. F_((s))|Fe^(2+)(0.001 M) ||H^(+)(1M) | H_(2(g)) ( 1 bar ) | Pt_((s)) is

    A
    `E_"cell"=E_"cell"^@-"0.591"/2 "log" ([Fe^(2+)][H^+]^2)/([Fe][H_2])`
    B
    `E_"cell"=E_"cell"^@-"0.591"/2 "log" ([Fe][H^+]^2)/([Fe^(2+)][H_2])`
    C
    `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([Fe^(2+)][H_2])/([Fe][H^+]^2)`
    D
    `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([Fe][H_2])/([Fe^(2+)][H^+]^2)`
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    The electrode potential of electrode M(s) rightarrow M^(n+) (aq) (2M) + ne^(-) at 298 K is E_(1). When temperature is doubled and concentration is made half, then the electrode potential becomes E_(2) . Which of the following represents the correct relationship between E_(1) and E_(2) ?