Problem: Given the following two half reactions and their potentials, which net reaction is spontaneous? Zn2+(aq) + 2 e– → Zn(s)      E°cell = –0.763 VNi2+(aq) + 2 e– → Ni(s)        E°cell = –0.25 VA. Zn2+(aq) + Ni(s) → Zn(s) + Ni2+(aq)B. Ni2+(aq) + Zn(s) → Ni(s) + Zn2+(aq)C. Ni(s) + Zn(s) → Zn2+(aq) + Ni2+(aq)D. Zn2+(aq) + Ni2+(aq) → Ni(s) + Zn(s) E. Zn2+(aq) + Zn(s) → Ni(s) + Ni2+(aq)

FREE Expert Solution

We’re being asked to determine the net reaction that is spontaneous.

We're given the following

Zn2+(aq) + 2 e → Zn(s)      E°cell = –0.763 V
Ni2+(aq) + 2 e → Ni(s)        E°cell = –0.25 V


Recall that the greater the E°cell of a redox reaction, the more likely the reaction will occur (more spontaneous reaction)

E°cell > 0 spontaneous in the forward direction
E°cell < 0 nonspontaneous in the forward direction


Determine the anode and cathode by comparing their E° values.

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Problem Details

Given the following two half reactions and their potentials, which net reaction is spontaneous? 

Zn2+(aq) + 2 e → Zn(s)      E°cell = –0.763 V
Ni2+(aq) + 2 e → Ni(s)        E°cell = –0.25 V


A. Zn2+(aq) + Ni(s) → Zn(s) + Ni2+(aq)

B. Ni2+(aq) + Zn(s) → Ni(s) + Zn2+(aq)

C. Ni(s) + Zn(s) → Zn2+(aq) + Ni2+(aq)

D. Zn2+(aq) + Ni2+(aq) → Ni(s) + Zn(s) 

E. Zn2+(aq) + Zn(s) → Ni(s) + Ni2+(aq)

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Our tutors have indicated that to solve this problem you will need to apply the Cell Potential concept. If you need more Cell Potential practice, you can also practice Cell Potential practice problems.