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The potential of the cell V(s)|V^(3+)(...

The potential of the cell
`V(s)|V^(3+)(aq., 0.0011 M) ||Ni^(2+)(aq., 0.24 M)||Ni(s)` is

A

`0.50 V`

B

`0.40 V`

C

`0.70 V`

D

`0.80 V`

Text Solution

Verified by Experts

The correct Answer is:
C
The two half reactions and their standard potential are
`{:(V^(3+)(aq.)+3e^(-)hArr V(s)E_(V^(3+)//V)^(@)=-0.89 V),(Ni^(2+)(aq.)+2e^(-) hArr Ni(s) E_(Ni^(2+)//Ni)^(@) = -0.23 V):}`
The `V^(3+)//V` couple is the anode thus `E^(@)` for the overall reaction is `E_("cell")^(@) = E_(R)^(@)- E_(L)^(@)`
`= (-0.23 V)-(-0.89 V)`
`= 0.66 V`
To obtain the overlal racito, we reverse the vanadium half-reaction. We multiply it by two and the nikel half-reaction by three.The number of elctrons that cancel when the two half-reactions are combined is six, so `n=6`. The cell reaction is
`2V(s)+3Ni^(2+)(aq.)hArr 2V^(3+) (aq.)+3Ni(s)`
The expression for `Q` for the overall reaction is
`Q = (C_(V3+)^(2))/(C_(Ni2+)^(3))`
Using the calcualted value of `E_("cell")^(@)` and the given data on concentration, the potential of the cell can be calculated with the help of the Nernst equation:
`E_("cell") = E_("cell")^(@)-(0.0592 V)/(6)log Q`
`= (0.66 V)- ((0.0592 V)/(6)) "log" (0.0011)^(2)/(0.24)^(3)`
`= 0.70 V`
The values of `E_("cell")^(@)` and `E_("cell")` tell us a great deal about the bahaviour of the chemical systeam that corresponds to the cell reaction. The large postivie value of `E_("cell")^(@)` means that the reaction has a large equiblrium constant and proceeds substanitially to completion as written.
Since `E_("cell")` is even larger than `E_("cell")^(@)`, the systeam is further from equilibrium than it would be at standard concentrations.
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