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Determine the electromotive force of the...

Determine the electromotive force of the following galvanic cell at `25^(@)C`. Given : `E_(Cd^(2+)|Cd)^(@)=-0.402V`
`Cd|Cd^(2+)(0.02M)||H^(+)(0.2M)|H_(2)("0.5 atm")|Pt`

Text Solution

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`{:("Anode reaction: "Cd(s)" "rarr" "Cd^(2+)(aq)+2e),("Cathode reaction : "2H^(+)(aq)+2e" "rarr" "H_(2)(g)),(bar(" Cell reaction: "Cd(s)+2H^(+)(aq)" "rarr" "Cd^(2+)(aq)+H_(2)(g)" ")):}`
As 2 mol of electrons are involved in the cell reaction, n = 2. The Nernst equation for the cell reaction is,
`E_("cell")=E_("cell")^(@)-(0.059)/(n)log.([Cd^(2+)]xxp_(H_(2)))/([Cd][H^(+)]^(2))`
`"or, "E_("cell")=E_("cell")^(@)-(0.059)/(2)log.([Cd^(2+)]xxp_(H_(2)))/([H^(+)]^(2))" "(because [Cd]=1)`
`E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)=0.0-(-0.402)=+0.402V`
Putting `[Cd^(2+)]=0.02M, [H^(+)]=0.2M and p_(H_(2))=0.5 atm`
and `E_("cell")^(@)=0.402V` in the Nernst equation gives
`E_("cell")=(0.059)/(2)log.(0.02xx0.5)/((0.2)^(2))=0.419V`
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