Question 19.8: Calculating Ecell under Nonstandard Conditions Determine the...
Calculating E_{cell} under Nonstandard Conditions
Determine the cell potential for an electrochemical cell based on the following two half-reactions:
Oxidation: Cu (s) \longrightarrow Cu ^{2+}(a q, 0.010 M )+2 e ^{-}
Reduction: MnO _4^{-}(a q, 2.0 M )+4 H ^{+}(a q, 1.0 M )+3 e ^{-} \longrightarrow MnO _2(s)+2 H _2 O (l)
| SORT You are given the half-reactions for a redox reaction and the concentrations of the aqueous reactants and products. You are asked to find the cell potential. | GIVEN: [MnO _4^{-}] = 2.0 M; [H ^{+}] = 1.0 M; [Cu ^{2+}] = 0.010 M FIND: E_{cell} |
| STRATEGIZE Use the tabulated values of electrode potentials to calculate E_{\text {cell }}^{\circ} .
Then use Equation 19.9 to calculate E_{\text {cell }}. E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0592 V }{n} \log Q [19.9] |
CONCEPTUAL PLAN
E_{\text {an }}^{\circ},E_{\text {cat }}^{\circ}\longrightarrow E_{\text {cell }}^{\circ}
E°_{\text {cell }}, [MnO _4^{-}] , [H ^{+}], [Cu ^{2+}] \longrightarrow E_{\text {cell }}
E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0592 V }{n} \log Q |
| SOLVE Write the oxidation and reduction halfreactions, multiplying by the appropriate coefcients to cancel the electrons.
Find the standard electrode potentials for each halfreaction. Find E_{\text {cell }}^{\circ} . |
SOLUTION
Oxidation (Anode): 3[Cu (s) \longrightarrow Cu ^{2+}(aq) + \cancel{2e ^{-}} ] E° = 0.34 V Reduction (Cathode): 2[MnO _4^{-}(a q)+4 H ^{+}(a q)+\cancel{3 e ^{-}} \longrightarrow MnO _2(s)+2 H _2 O (l)] E° = 1.68 V \\ \overline{ 3 Cu (s)+ 2 MnO _4^{-}(a q)+8 H ^{+}(aq) \longrightarrow 3 Cu ^{2+}(a q)+ 2 MnO _2(s)+4 H_2O (l)}\\ E^{\circ}_{cell} = E^{\circ}_{cat} – E^{\circ}_{an} = 1.34 V |
| Calculate E_{\text {cell }} from E_{\text {cell }}^{\circ} .
The value of n (the number of moles of electrons) corresponds to the number of electrons (six in this case) canceled in the half-reactions. Determine Q based on the overall balanced equation and the given concentrations of |
E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0592 V }{n} \log Q
=E_{\text {cell }}^{\circ}-\frac{0.0592 V }{n} \log \frac{\left[ Cu ^{2+}\right]^3}{\left[ MnO _4^{-}\right]^2\left[ H ^{+}\right]^8}
=1.34 V -\frac{0.0592 V }{6} \log \frac{(0.010)^{2} }{(2.0)^{2}(1.0)^{8}} = 1.34 V – (–0.065 V) = 1.41 V |
| CHECK The answer has the correct units (V). The value of E_{\text {cell }} is larger than E_{\text {cell }}^{\circ} , as expected based on Le Châtelier’s principle because one of the aqueous reactants has a concentration greater than standard conditions and the one aqueous product has a concentration less than standard conditions. Therefore, the reaction has a greater tendency to proceed toward products and a greater cell potential. | |