Question

Using these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)/Cu(s) -0.40 V -0.76 V ‑0.25 V +0.34 V Calculate the standard cell potential for the cell whose reaction is Ni2+(aq) + Zn(s) →Zn2+(aq)+ Ni(s)

Answer 1

The standard cell potential for the given reaction is 0.51 V.

Step-by-step explanation:

The standard cell potential for the cell whose reaction is Ni2+(aq) + Zn(s) → Zn2+(aq) + Ni(s) can be calculated using the standard reduction potentials provided.

We need to identify the reduction half-reactions for Ni2+ and Zn2+.

The reduction half-reaction for Ni2+(aq) is Ni2+(aq) + 2e- → Ni(s) with a standard reduction potential of -0.25 V.

The reduction half-reaction for Zn2+(aq) is Zn2+(aq) + 2e- → Zn(s) with a standard reduction potential of -0.76 V.

To calculate the standard cell potential, we subtract the reduction potential of the anode (Zn/Zn2+) from the reduction potential of the cathode (Ni/Ni2+).

Therefore, the standard cell potential is (-0.25 V) - (-0.76 V) = 0.51 V.

Answer 2

Answer: The standard cell potential for the cell is +0.51 V

Step-by-step explanation:

Given :
`E^0_(Ni^(2+)/Ni)=-0.25V`

`E^0_(Zn^(2+)/Zn)=-0.76V`

The given reaction is:

`Ni^(2+)(aq)+Zn(s)\rightarrow Zn^(2+)(aq)+Ni(s)`

As nickel is undergoing reduction, it acts as cathode and Zinc is undergoing oxidation, so it acts as anode.

`E^0_(cell)=E^0_(cathode)-E^0_(anode)`

where both
`E^0` are standard reduction potentials.

Thus putting the values we get:

`E^0_(cell)=-0.25-(-0.76)`

`E^0_(cell)=0.51V`

Thus the standard cell potential for the cell is +0.51 V