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Calculate the standard potential for the reaction, `Hg_(2)Cl_(2)+Cl_(2) rarr 2Hg^(2+)+4Cl^(-)` Given : `{:(Hg_(2)Cl_(2)+2e^(-) rarr 2Hg+2Cl^(-),,E^(@)=0.270" volt"),(Hg_(2)^(2+) rarr 2Hg^(2+)+2e^(-),,E^(@)=-0.92" volt"),(2Hg rarr Hg_(2)^(2+)+2e^(-),,E^(@)=-0.79"volt"),(Cl_(2)+2e^(-) rarr 2Cl^(-),,E^(@)=1.36" volt"):}` |
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Answer» Correct Answer - `-0.08` volt First determine `E^(@)` for `Hg_(2)Cl_(2) rarr 2Hg^(2+) +2Cl^(-) +2e^(-)` electrode. It comes to -1.44 volt. This electrode is now coupled with `Cl_(2)+2e^(-) rarr 2Cl^(-)` electrode. |
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