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The potential of the cell `V(s)|V^(3+)(aq., 0.0011 M) ||Ni^(2+)(aq., 0.24 M)||Ni(s)` isA. `0.50 V`B. `0.40 V`C. `0.70 V`D. `0.80 V` |
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Answer» Correct Answer - 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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