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REDOX & Electrochemistry - LSU Chemistry

REDOX & Electrochemistry - LSU Chemistry

REDOX

REDOX 30Example: Calculate the potential for the Fe 3+ /Fe 2+electrode when the concentration of Fe 2+ is fivetimes greater than that of Fe 3+ .Look up potential in a half cell table:Fe 3+ + e- Fe 2+ Eº = +0.771 VThe Fe 2+ concentration is five times greater thanthat of Fe 3+ , so the Q expression is:p[products] [Fe2+]Q =r= = 53+[reactants] [Fe ]Assume room temp (25ºC = 298K) and solve usingthe Nernst equation:EEE0.0592= E− log Qn0.0592= 0.771 − log(5)1= 0.771 − (0.0592)(0.7) = 0.730 VProblem: The potential for this is lower than Eºwhere both concentrations are 1 M. This shouldmake sense based on Le Chatelier’s principle.Why?

REDOX 31Problem: A cell is constructed at 25ºC as follows. Onehalf-cell consists of Cu 2+ /Cu, [Cu 2+ ] = 0.4 M. The otherhalf-cell involves Zn 2+ /Zn with [Zn 2+ ] = 0.4 M. Applythe Nernst equation to the overall cell reaction todetermine the cell potential.Problem: Consider the following half-cell rxn:Pb(s) + SO 4 2− (aq) PbSO 4 (s) + 2e- Eº = +0.13 VCan the Nernst equation help us figure out how tochange concentrations to increase the cell potential?Why?

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