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The reduction potential diagram (Latimer diagram) for Cu is acid solution Calculate x. Does Cu^(+) disproportionate in the solution? |
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Answer» Solution :`Cu^(2+),e to Cu^+ , E^@=0.15 , Delta G^@=n F E^@=-1 times 0.15 F =-0.15 F` `Cu^+ + e to Cu: E^@= 0.50, Delta G^@=-1 times 0.50 F=-0.50 F` On adding `Cu^(2)+2 e to Cu, Delta G^@=-0.65 F` `E_(Cu^(2+),Cu)^@=(Delta G^@)/(- n F)=(-0.65 F)/(-2F)=0.325 ` volt `x=0.325` volt Further `Cu^+ to Cu^(2+) +e, Delta G^@=-nFE^@=-1 times (-0.15)F=0.15 F` `Cu^+ +e to Cu, Delta G^@=-nFE^@=-1 times 0.50 times F=-0.50 F` `=0.0591/n log {[Ag^+]_(righ t)/[Ag^+]_(l eft)}` `=0.0591/1 log""0.1/0.001=0.0591 log 100` `=0.1182` volt Since `E_(CELL)` is positive, the right electrode will act as CATHODE (where REDUCTION occurs) and the left electrode will act as anode (where oxidation occurs) The electrons will thus travel from left to right in external circuit. |
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