Cell Potential Question

Question

Calculate the Eocell for the redox reaction shown below in volts:

Zn(s)+Ni+2(aq)→Zn+2(aq)+Ni(s)

Answer 1

The question asks to find the cell potential in volts for the following RXN

Zn + Ni+2>>>>Zn+2 + Ni

In order to do this you must first determine in this RXN what has been oxidized (lost electrons) and what has been reduced (gained electron). To determine this you use a table of standard cell reduction potentials. The more positive value is what has been reduced, the more negative value is what has been oxidized.

With that you re-write the above equation if needed.

Here are the reduction potentials: Zn+2 + 2e -0.76 V

Ni+2 + 2e -0.257V

Zn is what is being oxidized and Ni is what is being reduced.

the half cell RXN's are now: Zn >>> Zn+2 + 2e (oxidized)

Ni+2 + 2e>>> Ni (reduced)

in a Galvanic cell, oxidation takes place at the Anode and Reduction takes place at the Cathode.

To find the standard cell Voltage you use this formula: E0 (v) = reduction (v) - oxidation {v)

Therefore: E0 = -0.257 - (-0.76) = 0.503 volts

:

Answer 2

The cell can be written:

Zn(s) | Zn2+ (aq) || Ni+2 (aq) | Ni(s)

E^o_cell = Reduction potential at cathode - reduction potential at anode.

From the internet:

Reduction potential at cathode = -0.26 V

reduction potential at anode = - 0.76 V

E^o_cell = - 0.26 V- ( - 0.76 V) = + 0.5 V