#sf(Zn^(2+)+2erightleftharpoonsZncolor(white)(xxxxx)" E^@=-0.76color(white)(x)V)#
#sf(Ni^(2+)+2erightleftharpoonsNicolor(white)(xxxxxx)E^@=-0.25color(white)(x)V)#
Since the zinc 1/2 cell has the more -ve value this will go right to left so the cell reaction is:
#sf(Zn+Ni^(2+)rarrZn^(2+)+Ni)#
To find #sf(E_(cell)^@)# you subtract the least +ve value from the most +ve:
#sf(E_(cell)^@=-0.25-(-0.76)=+0.51color(white)(x)V)#
Consider the initial (I) and final (F) conditions based on #sf("mol/l")#:
#" "sf(Zn+Ni^(2+)" "rarr" "Zn^(2+)+Ni)#
I #sf(color(white)(xxxxx)1.60color(white)(xxxxxxx)0.130)#
C #sf(color(white)(xxxx)-xcolor(white)(xxxxxxx)+x)#
F #sf(color(white)(xxx)(1.60-x)color(white)(xx)(0.130+x))#
The concentration of #sf(Ni^(2+))# falls and the concentration of #sf(Zn^(2+))# increases as current is drawn from the cell.
The reaction quotient when #sf(E_(cell)^@)# has fallen to + 0.45 V is given by:
#sf(Q=([Zn^(2+)])/([Ni^(2+)])=((0.130+x))/((1.60-x))#
To find the value of #sf(Q)# we use The Nernst Equation which can be simplified to:
#sf(E_(cell)=E_(cell)^@-0.0591/zlogQ)#
(At #sf(25^@C)#)
#sf(z)# is the no. of moles of electrons transferred which, in this case, is 2.
#:.##sf(0.45=0.51-0.0591/2logQ)#
#sf(logQ=(0.06xx2)/0.0591=2.3045)#
#sf(Q=107.26)#
#:.##sf(((0.130+x))/((1.60-x))=107.26)#
#sf(0.130+x=107.26(1.60-x))#
#sf(0.130+x=171.623-107.26x)#
#sf(108.26x=171.49)#
#sf(x=171.49/108.26=1.584)#
#:.##sf([Zn^(2+)]=0.130+1.584=1.714color(white)(x)"mol/l")#
#sf([Ni^(2+)]=1.60-1.584=0.016color(white)(x)"mol/l")#
As expected, the concentration of #sf(Zn^(2+))# ions has increased and the concentration of #sf(Ni^(2+))# ions has fallen as work is done by the cell.
