Ni2+ + 2e- →Ni (E°= -0.28 V)
Zn2+ + 2e- →Zn (E°= -0.76 V)
I need to write the overall reaction and cell potential. I think I sort of understand how to do it but am not sure... do I just add the two voltages to get the cell potential? So -104?
Reaction+ cell potential for redox reaction in galvanic cell?
2018-01-23 3:44 pm
回答 (1)
2018-01-23 4:29 pm
Ni²⁺(aq) + 2e⁻ → Ni(s) …… E° = -0.28 V
Zn²⁺(aq) + 2e⁻ → Zn(s) …… E° = -0.76 V
As the second half reaction has higher reduction potential, Zn is a stronger reducing agent than Ni. Therefore, Zn will undergo reduction and Ni²⁺ will undergo oxidation.
Zn(s) + Ni²⁺(aq) → Zn²⁺(aq) + Ni(s) …… E°(cell)
E°(cell) = E°(Red) - E°(Oxi) = (-0.28 V) - (-0.76 V) = +0.48 V
Zn²⁺(aq) + 2e⁻ → Zn(s) …… E° = -0.76 V
As the second half reaction has higher reduction potential, Zn is a stronger reducing agent than Ni. Therefore, Zn will undergo reduction and Ni²⁺ will undergo oxidation.
Zn(s) + Ni²⁺(aq) → Zn²⁺(aq) + Ni(s) …… E°(cell)
E°(cell) = E°(Red) - E°(Oxi) = (-0.28 V) - (-0.76 V) = +0.48 V
收錄日期: 2021-04-24 00:56:25
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