Question #4a3ac

1 Answer
May 15, 2015

Answer:

#E^(@)=+0.37"V"#

#DeltaG^(@)=-35.7"kJ"#

#K_c=1.82xx10^(6)"mol"^(-1)."l"#

Explanation:

Start with the #E^(@)# values, listing them in increasing order:

#Cu_((aq))^(2+)+erightleftharpoonsCu_((aq))^(+)# #E^(@)=+0.153"V"# #color(red)((1))#

#Cu_((aq))^(+)+erightleftharpoonsCu_((s))# #E^(@)=+0.52"V"# #color(red)((2))#

To work out #E_((cell))^(@)# subtract the least +ve value from the most +ve.:

#E_((cell))^(@)=+0.52-0.153=+0.37"V"#

This means #color(red)((2))# will be driven left to right and #color(red)((1))# will be driven right to left#rArr#

#2Cu_((aq))^(+)rarrCu_((aq))^(2+)+Cu_((s))#

#DeltaG^(@)=-nFE_((cell))^(@)#

#= -1xx9.65xx10^(4)xx0.37=-35705"J"#

#=-35.7"kJ"#

#DeltaG^(@)=-RTlnK_c#

#lnK_c=(-DeltaG)/(RT)#

#lnK_c=(35.7xx10^(3))/(8.31xx298)=14.416#

From which:

#K_c=1.82xx10^(6)"mol"^(-1)."l"#

This is an example of a disproportionation reaction where a species, in this case copper(1), is simultaneously oxidised and reduced.