A cell with EMF = +1.30 V is made from two electrodes. The half-equations for the two electrodes are shown:
Positive electrode:
NiO(OH)(s)+H2O(l)+e−→Ni(OH)2(s)+OH−(aq) \text{NiO(OH)}(\text{s}) + \text{H}_2\text{O}(\text{l}) + \text{e}^- \rightarrow \text{Ni(OH)}_2(\text{s}) + \text{OH}^-(\text{aq}) NiO(OH)(s)+H2O(l)+e−→Ni(OH)2(s)+OH−(aq)Negative electrode:
Cd(OH)2(s)+2e−→Cd(s)+2OH−(aq)E⊖=−0.81 V \text{Cd(OH)}_2(\text{s}) + 2\text{e}^- \rightarrow \text{Cd}(\text{s}) + 2\text{OH}^-(\text{aq}) \quad E^\ominus = -0.81\text{ V} Cd(OH)2(s)+2e−→Cd(s)+2OH−(aq)E⊖=−0.81 VWhat is the standard electrode potential of the NiO(OH)/Ni(OH)2\text{NiO(OH)} / \text{Ni(OH)}_2NiO(OH)/Ni(OH)2 electrode?
−2.11 V-2.11\text{ V}−2.11 V
−0.49 V-0.49\text{ V}−0.49 V
+0.49 V+0.49\text{ V}+0.49 V
+2.11 V+2.11\text{ V}+2.11 V
