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.
