Write Net cell notation; calculate cell potential.?

(a) Identify the cathode and the anode

The standard convention for cell notation puts the anode on the left and the cathode on the right.

Thus, the anode is #"Pb(s)"# and the cathode is #"Ni(s)"#.

(b) Write the net cell reaction

#color(white)(mmmmlmmmmmmmmmmmmmmmmmmmmll)ul(E^@"/V")#
#"Anode": color(white)(m)"Pb(s)" → "Pb"^"2+""(aq)" + 2"e"^"-";color(white)(mmmmmmml) "+0.126"#
#"Cathode": ul("Ni"^"2+""(aq)" + 2"e"^"-" → "Ni(s)"; color(white)(mmmmmmll))color(white)(mll)ul("-0.25")#
#"Cell":color(white)(mm)"Pb(s)" + "Ni"^"2+""(aq)" → "Pb"^"2+""(aq)" + "Ni(s)"; color(white)(m)"-0.124"#

(c) Calculate the net cell potential

#E_text(cell)^@ = "-0.124 V"#

The cell reaction is not spontaneous as written.

Here's my approach...

From standard reduction potential tables:

#"Pb"^(2+)(aq) + 2e^(-) -> "Pb"(s)#, #E_(red)^@ = -"0.13 V"#
#"Ni"^(2+)(aq) + 2e^(-) -> "Ni"(s)#, #E_(red)^@ = -"0.25 V"#

#E_(red)^@# for #"Pb"# is less negative, so #"Pb"^(2+)# is reduced more easily.

Therefore, the cathode involves #"Pb"^(2+)# and #"Pb"(s)#, and the anode involves #"Ni"^(2+)# and #"Ni"(s)#.

We then have:

#"Pb"^(2+)(aq) + cancel(2e^(-)) -> "Pb"(s)#, #E_(red)^@ = -"0.13 V"#
#ul("Ni"(s) -> "Ni"^(2+)(aq) + cancel(2e^(-)))#, #E_(o x)^@ = +"0.25 V"#
#color(blue)("Pb"^(2+)(aq) + "Ni"(s) -> "Ni"^(2+)(aq) + "Pb"(s))#

#color(blue)(E_(cell)^@) = E_(red)^@ + E_(o x)^@#

#= -"0.13 V" + "0.25 V" = color(blue)(+"0.12 V")#

Or, if BOTH are reduction potentials:

#color(blue)(E_(cell)^@) = E_(cathode)^@ - E_(anode)^@#

#= -"0.13 V" - (-"0.25 V") = color(blue)(+"0.12 V")#