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Find the the reducation potential of the half-cell `Pt(s)|Cu^(2+)(aq., 0.22 M), Cu^(+)(aq., 0.043 M)`A. `0.20 V`B. `-0.20 V`C. `0.30 V`D. `-0.30 V` |
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Answer» Correct Answer - A The half-cell reaction is `Cu^(2+)(aq.)+e^(-)hArr Cu^(+)(aq.)` and its `E_(cu^(2+)//Cu)^(@)` is `0.158 V` The value of `n`, the number of electrons transferred, is clearly `1`, for this half-reaction. The expression for `Q` is `C_(Cu^(+)//C_Cu^(2+))` Substituting these data into the Nernst equaltion. `E= E^(@) - (0.0592)/(n) log Q` `E_(Cu^(2+)//Cu^(+)) = E_(Cu^(2+)//Cu^(+))-(0.0592 V)/(n) "log" (C_(Cu^(+)))/(C_(Cu^(++)))` ` (0.158 v) - (0.0592 V)/(1) "log" (0.043)/(0.22)` `= 0.20 V` This answer is consistent with the chemistry of this system. The positive value of `E_("cell")` tells us that `Cu^(+)` (aq.) is stable relative to `Cu^(2+)(aq.)` and will predominte at equibirium. In this half-cell, the systeam is even further from equilibrium than in the standard state because the `[Cu^(2+)]` is sibstanitially larger than the `[Cu^(+)]`. Therefore, `E` is greater than `E^(@)`. |
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