Question #18173

1 Answer
Jan 6, 2018

Answer:

#sf(E^@=+1.29color(white)(x)V)#

Explanation:

You haven't given the relevant #sf(E^@)# values so I looked them up:

#sf(Au^(3+)+3erightleftharpoonsAucolor(white)(xxxxxxxx)E^@=+1.42color(white)(x)V)#

#sf(Au^(+)+erightleftharpoonsAucolor(white)(xxxxxxxxxx)E^(@)=+1.68color(white)(x)V)#

#sf(Au^(3+)+2erightleftharpoonsAu^(+)color(white)(xxxxxxx)##sf(E^@# is unknown

We need to convert these to free energy values and then apply the conservation of energy.

#sf(DeltaG^@=-nFE^@)#

#sf(F)# is the Faraday Constant

#sf(n)# is the no. of moles of electrons transferred.

Let #sf(E^@)# be the value we want to find.

We can say that:

#sf(DeltaG^@[Au^(3+)rarrAu]=DeltaG^@[Au^(3+)rarrAu^(+)]+DeltaG^@[Au^(+)rarrAu])#

#:.##sf(-3cancel(F)xx1.42=-2cancel(F)E^(@)-cancel(F)xx1.68)#

#:.##sf(-2E^@=-4.26+1.68=-2.58color(white)(x)V)#

#sf(E^@=+2.58/2=+1.29color(white)(x)V)#