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To find the standard potential of M^(3+)...

To find the standard potential of `M^(3+)//M` electrode , the following cell is constituted : `Pt|M|M^(3+((.001moLL^(-1)|| Ag ^(+) ((0.01moLL^(-1))|` Ag. The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction `M^(3+)+ 3e^(-) to M` at 298 K will be :
(Given `E_(Ag^(+)//Ag^(@)` at 298 K = 0.80 Volt)

Text Solution

Verified by Experts

The cell reaction is
`M+3Ag^(+) rarr 3Ag+M^(3+)`
Applying Nernst equation,
`E_(cell)=E_(cell)^(@)-0.0591/3"log" ([M^(3+)])/([Ag^(+)]^(3))`
`0.42=E_(cell)^(@)-0.0591/3"log"((0.0018))/((0.01)^(3))=E_(cell)^(@)-0.064`
`E_(cell)^(@)=(0.42+0.064)=0.484` volt
`E_(cell)^(@)=E_("Cathode")^(@)-E_("Anode")^(@)`
or `E_("Anode")^(@)=E_("Cathode")^(@)-E_(Cell)^(@)`
`=(0.80-0.484)=0.32` volt
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