Determine the cell potential for Ni(s) + Fe²⁺(aq) → Ni²⁺(aq) + Fe(s) where [Ni²⁺] = 0.60 M and [Fe²⁺] = 0.0030 M using the following standard reduction potentials. Ni²⁺(aq) + 2e⁻ → Ni(s) E° = -0.25 V and Fe²⁺(aq) + 2e⁻ → Fe(s) E° = -0.44 V.