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Calculating the Gibbs Energy change from electrode Potentials: Using standard electrode potentials, clacu-late the standard Gibbs energy change at `25^(@)C` for the following cell reaction. Strategy: To calculate `DeltaG_("cell")^(@)`, we use the equation `DeltaG_("cell")^(@) = -nFE_("cell")^(@)`, where `E_("cell")^(@)` is obtained by using a table of standard potentails and `n` can be inferred from the balanced chemical equation. The cell reaction equals the sum of the half-cell reactions after they have been multiplied by factors so that the electrons cancel in the summation. Note that `n` is the number of electrons involved in each half-reaction. |
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Answer» The half-reactions, corresponding half-cell potentials, and their sums are displayed below `{:(Zn(s)hArrZn^(2+)(aq., 1M)+2e^(-)" "E_(Zn//Zn^(2+))^(@)=0.76V),(2Ag^(+)(aq., 1M)=2e^(-)hArr 2Ag(s)" "E_(Ag^(+)//Ag)^(@) = 0.80 V),(bar(Zn(s)+2Ag^(+)(aq., 1M)hArr Zn^(2+)(aq.,1M)+2Ag(s)E_("cell")^(@)=1.56V)):}` Note that each half-reaction involves two electrons, hence, `n = 2` Also, `E_("cell")^(@) = 1.56 V`, and the Faraday constant, `F`, is `9.65 xx 10^(4) C`. Therefore, the standard Gibbs energy change is `DeltaE_("cell")^(@) = -nFE_("cell")^(@)` `-(2 "mole"^(-))(96,500 C//mol e^(-)) (1.56 V) ((1 J)/(1 C. V))` ` = -3.01 xx 10^(5)J`, Recell that (Coulombs) `xx` (volts) = joules `= -301 kJ (1 kJ = 10^(3)J` Short cut : `F` is about `10^(5)C//mol e^(-)` and `E^(@)` is about `1.5 V`, so `DeltaG^(@) = -nFE^(@)` is approximately `- (2 mol e^(-))(10^(5)C//mol e^(-)(1.5 V) = -3.0 xx 10^(5) J`, or `300 kJ`. |
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