anode reaction:
oxidation takes place
Sn(s) -------------------------> Sn+2 (aq) + 2e-Â Â
,  E0Sn+2/Sn = - 0.15 V
cathode reaction
: reduction takes palce
Ni+2(aq) + 2e- -----------------------------> Ni(s) ,
E0Ni+2/Ni = -0.25V
--------------------------------------------------------------------------------
net reaction: Ni2+ (aq) + Sn (s) → Ni (s) +
Sn2+ (aq)
E0cell= E0cathode-
E0anode
E0cell= E0Ni+2/Ni -
E0Sn+2/Sn
         = -0.25 -
(-0.15)
         = -0.10
V
detla G0 = -nF Ecell0
            Â
= - 2x 96, 485 x (-0.10)
            Â
= 19297 J
             Â
= 19.3 kJ
ΔG° = 19.3 kJ
