To find #E_(cell)# when we do not have standard conditions we use the Nernst Equation:
#E_(cell)=E_(cell)^@-(RT)/(nF)lnQ#
At 298K this simplifies to:
#E_(cell)=E_(cell)^@-(0.05916)/(n)logQ#
#n# is the number of moles of electrons transferred.
#Q# is the reaction quotient.
To find #E_(cell)^@# and the cell reaction list the #E^@# values most -ve to most +ve :
#" "E^@"(V)"#
#stackrelcolor(blue)(larr)color(white)(xxxxxxxxxxxxxxxxxxxxxxxxx)#
#Sn^(2+)+2erightleftharpoonsSn" "-0.136#
#TiO^(2+)+e+2H^+rightleftharpoonsTi^(3+)+H_2O" "+0.1#
#stackrelcolor(red)(rarr)color(white)(xxxxxxxxxxxxxxxxxxxxxxxxx)#
The most +ve 1/2 cell is the one that takes in the electrons so you can see that #TiO^(2+)# will oxidise #Sn# to #Sn^(2+)#
The cell reaction is therefore:
#Sn_((s))+2TiO_((aq))^(2+)+4H_((aq))^(+)rarrSn_((aq))^(2+)+2Ti_((aq))^(3+)+2H_2O_((l))#
The reaction quotient is therefore given by:
#Q=([Sn^(2+)][Ti^(3+)]^2)/([TiO^(2+)]^(2)[H^+]^4)#
To find #E_(cell)^@# subtract the least +ve 1/2 cell from the most +ve#rArr#
#E_(cell)^@=+0.1-(-0.136)=+0.236"V"#
Now insert the values into the Nernst Equation:
#E_(cell)=0.236-(0.05916)/(2)xxlog[((0.1)(0.1)^2)/((0.2)^2(0.05^2)]]#
#E_(cell)=0.236-(0.05916)/(2)log[400]#
#E_(cell)=0.236-[0.02958xx(2.6)]#
#E_(cell)=0.236-0.077#
#E_(cell)=+0.16"V"#
