J.R. S. answered 11/28/23
Ph.D. University Professor with 10+ years Tutoring Experience
NiO(OH)(s) + H2O (l) + e- Ni(OH)2 (s)+ OH- (aq) ... reduction half reaction (cathode)
Cd (s) + 2OH- (aq) Cd (OH)2 + 2e- ... oxidation half reaction (anode)
Multiply reduction half reaction by 2 in order to equalize electrons, then add the 2 reactions together:
2NiO(OH)(s) + 2H2O (l)+ 2e- 2Ni(OH)2 (s)+ 2OH-(aq)
Cd (s) + 2OH- (aq) Cd (OH)2 + 2e-
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2NiO(OH)(s) + Cd(s) + 2H2O(l) 2Ni(OH)2 (s) + Cd (OH)2 ... BALANCED REDOX
If you want to Eºcell, it is simply 1.32 V + 0.40 V = 1.72V