The standard oxidation potential potential of zinc is 0.76 volt and of silver is - 0.80 volt. Calculate the emf of the cell : `Zn|underset(0.25 M)(Zn(NO_(3))_(2))||underset(0.1 M)(AgNO_(3))|Ag` at `25^(@)C`.

The standard oxidation potential potential of zinc is 0.76 volt and of

1.	The standard oxidation potential potential of zinc is 0.76 volt and of silver is - 0.80 volt. Calculate the emf of the cell : `Zn\|underset(0.25 M)(Zn(NO_(3))_(2))\|\|underset(0.1 M)(AgNO_(3))\|Ag` at `25^(@)C`.
Answer» The cell reaction is `Zn+2Ag^(+) rarr 2Ag+Zn^(2+)` `E_(o x)^(@)` of `Zn =0.76` volt `E_("red")^(@)` of `Ag=0.80` volt `E_(cell)^(@)=E_(o x)^(@)` of `Zn+E_("red")^(@)` of `Ag=0.76+0.80=1.56` volt We know that, `E_(cell)=E_(cell)^(@)-0.0591/n"log" (["Products"])/(["Reactants"])` `=E_(cell)^(@)-0.0591/2"log" 0.25/(0.1xx0.1)` `=1.56-0.0591/2xx1.3979` `=(1.56-0.0413)` volt `=1.5187` volt Alternative method : First of all, the single electrode potentials of both the electrodes are determined on the basis of given concentrations. `E_("ox (Zinc)")=E_(o x)^(@)-0.0591/2log 0.25` `=0.76+0.0177=0.7777` volt `E_("red (Silver)")=E_("red")^(@)+0.0591/1 log 0.1` `=0.80-0.0591` `=0.7409` volt `E_("cell")=E_("ox (Zinc)")+E_("red (Silver)")` `0.7777+0.7409` `=1.5186` volt

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The standard oxidation potential potential of zinc is 0.76 volt and of silver is - 0.80 volt. Calculate the emf of the cell : `Zn|underset(0.25 M)(Zn(NO_(3))_(2))||underset(0.1 M)(AgNO_(3))|Ag` at `25^(@)C`.

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