InterviewSolution
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InterviewSolution
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The electrotrode potentials for `Cu^(2+)(aq.)+2e^(-) rarr Cu^(+)(aq.)` and `Cu^(+)(aq.)+e^(-) rarr Cu(s)` are `+0.15 V` respectively. The value of `E_(Cu^(2+))^(ɵ)//Cu` will be:A. `0.150 V`B. `0.500 V`C. `0.325 V`D. `0.650 V` |
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Answer» Correct Answer - C Many elements exist in a number of different oxidation states. The emf does not list the value of all the standard potentials of every possible half-reaction for these elements. But we can obtain these values, as long as each oxidation state of interest appers in a half-reaction in the series. To obtain this kind of information, we must combine half-reactions in the series to make a new half-reaction that is not in the series. The standard potential of the new half-reaction cannot bc found simply by a combination of the potentials of the original half-reactions. We could conver the `E^(o+)` of each half-reaction to the corresponding `DeltaG^(@)`, combine the `DeltaG^(@)` value to find the `DeltaG^(@)` back to `E^(@)`. But we shall use a different method. The `E^(@)` of each half-reaction is multiplied by the number of electrons in that half-reaction is multiplied by the number of electrons in that half-reaction. The resulting values arc combined. Finally, we divide the resulting potential by the number of electrons in the new half-reaction to get its potential. The procedure is sysbolized by `E_(T)^(@) = (n_(1)E_(1)^(@)+n_(2)E_(2)^(@)+.....)/(n_(T))` where `n_(1)` is the number of electrons in the first half-reaction and `E_(1)^(@)` is its standard potential, `n_(2)` is the number of electrons in the second half-reaction and `E_(2)^(@)` is its standard potential (and so on for as many half-reactions as are baing combined),and `n_(T)` is the number of electrons that appear in the resulting half-reaction. In using this relationship, be careful to make the sign of each `E^(@)` consistant with the way in which its half-reaction is combined to amke the new half-reaction. we being by writing the desired half-reaction. `Cu^(2+)(aq.)+2e^(-) rarr Cu(s)` Now we write half reactions that include these oxidation states of copper. These arc `Cu^(2+)(aq.)+e^(-) rarr Cu^(+)(aq.)," "E_(1)^(@) = +0.15 V` `Cu^(2+)(aq.)+e^(-) rarr Cu(s)," "E_(2)^(@) = +0.50 V` We see that the two equations should be added to produce the desired half-reaction. Thus `E_(T)^(@)=((1)(+0.15 V)+(1)(+.50 V))/(2)` `= 0.325 V` |
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