The Gibb's energy of equation-3 is -10615kj .
Given ,
1) Fe3+(aq) +e =Fe2+
2) Fe2+(aq) +2e = Fe(s)
3) Fe3+(aq) +3e = Fe (s)
the E cell of the equation-1 is 0.77 volts .
thus, [tex]delG^{0} =nFEcell = 1 \times96500\times0.77=74305 kj[/tex]
then the E cell of equation-2 is -0.44 volts .
thus ,[tex]delG^{0} =nFEcell =2\times 96500\times(-0.44)=-84920 kj[/tex]
by adding two equations 1 and 2 ,we get the equation 3 ,
thus , Gibbs energy of equation-3 is given by ,
[tex]delG^{0} = 74305 +(-84920) =-10615 kj[/tex]
Thus , the Gibbs energy of equation-3 is -10615kj .
Learn more about Gibb's energy here,
brainly.com/question/17310317
#SPJ4
