# Example hydrogen peroxide h2ow is possible product of

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example: hydrogen peroxide (H2Ow) is possible product of reduction of oxygen in acidicsolution: oO2(g) + 2H3O+(aq) + 2e- H2O2(l) + 2H2O(l), E3* = ?oIt can be further reduced to water: H2O(l) + 2H3O+(aq) + 2e- 4H2O(aq), E2* =1.77 VO2 –( ? = .688 V)--> H2O2oUse the half-potnetial given for the reduction of H2O2, togher with that given earlier, O2(g) + 3H2O+(aq) + 4e- 6H2O(l), E*1 = 1.229 VTo calculate the standard half-cell potential of O2 to H2O2 in acidic solution. ΔGhc = ΔG1 – ΔG2-nFEhc = -nFE1* + nFE2*-2(F)(Ehc) = -4(F)(E1) + 2(F)(E2)-Ehc = -2*E1 + E2Ehc = 2E1 – E2 = 2(1.229 V) – 1.77 V = .688 VThe coefficients from n values comes from number of electrons exchanged (the coefficients for the electrons corresponding to eachreaction oWrite a reduction potential diagram for O2, H2O2, and H2O. O2 .688 VH2O2 1.77 V(E2)H2O|_________1.228V (E1)______________|oIs H2O2 stable with respect to disproportionation in acidic solution?It will disproportionate because value is something? Smaller maybeoConcentration effects and the Nernst EquationApply thermodynamic principles to understand how concentration and pressure affect cell voltageNernst equation: Ecell = E*cell – (RT/nF)lnQYou can rewrite in log10 terms2/5/16
PracticeoFollowing galvanic cell constructed: Cr(s)|Cr3+(aq)||Cl2(g)|Cl-(aq)oWrite balanced chemical equations for the half-reactions at the anode and the cathode and for the overall cell reactionoCalculate ehte cell voltage, assuming that all reactants and products arein their standard statesoE*cell = Ecathode – EanodeCl2(g) + 2e- 2Cl-(aq), E* = 1.3583 V …1.3583 – (- .74 V) = 2.0983 VComputing cell votage from free energy changeoFigure out overall reaction to figure out how many electrons transferred in the overall reactionΔE* = ΔE1* = ΔE2* = 1.3583 V – (0.74 V)Adding and subtracting half-cell reactionsoTo compute half-cell voltage from other half-cell voltages, first compute the free energy change. Half-cell potentials are intensive functions. oExample: compute half-cell potential of: Cu2+(aq) + e- Cu+(aq) (reaction 3)Cu2+ Cu+ Cu\_______________/^Practice: calculate the half-cell potential of Au3+(aq) + 2e- Au+(aq)oCannot directly apply, so must use ΔG calculationoΔG of overall is Au3+ + Au+ Reduction potential diagrams and disproportionation oHalf-reactions of copper can be summarizedoCu+ is not stable, prefers to form Cu and Cu2+oDisproportionation: process in which single substance is both oxidized and reduced. Can occur if reduction immediately to right is lager than oen to the left. Driving force is difference in two reduction potentials. Since the value is greater, we know that it’s more likely to form CuConcentration effects and the Nernst EquationoQ = pressures in atm (or bar) (standard state (products over reactants) raised to their respective coefficientso–nFEcell = -nFE*cell + RTlnQoΔG = ΔG* + RTlnQ (from this equation)Nernst Equation: oEcell = E*cell – (RT/nF)lnQoOr Ecell = E*cell - .(0592 V/n)logQ (at 25*C)Example: oA galvanic cell is constructed that carries out the reactionPb2+(aq) + 2Cr2+(aq) Pb(s) + 2Cr3+(aq)oCalculate the cell potential at 25*C under each of the following conditions: Standard conditionsE*cell = E*cathode – E*anode
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