The reduction potential diagram (Latimer diagram) for Cu is acid solution

Calculate x. Does `Cu^(+)` disproportionate in the 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.