The reduction potential diagram for `Cu` in acid solution is :

Calculate `X`. Does `Cu^(+)` disproportionate in solution ?

Explanation

`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.