Question

Construct a cell from Ni2+ | Ni and Cu2+ | Cu Cu half cells. Write the cell reaction and calculate `E_("cell")^0`   
`(E_("Ni")^0 = - 0.236  V and E_("Cu")^0 = + 0.337  V)`

Answer 1

The standard potential of Cu₂₊ | Cu half cell is greater than that of Ni²⁺ | Ni half cell. Therefore, Cu²⁺ | Cu half cell is cathode (RHS) and Ni²⁺ | Ni half cell is anode (LHS). The cell is Ni(s) | Ni²⁺ (1M) || Cu²⁺ (1 M) | Cu.

Electrode reactions and net cell reaction:
Oxidation at anode     : Ni(s)→ Ni²⁺ (1 M) + 2e^-
Reduction at cathode : Cu²⁺ (1 M) + 2e^- → Cu(s)
______________________________________________________
Net cell reaction         : Ni(s) + Cu²⁺ (1 M) → Ni²⁺ (1 M) + Cu(s)

`E_("cell")^0 = E_("cathode")^0 - E_("anode")^0`

= 0.337 - (- 0.236) = 0.573 V