Problem 15 Excess of zinc reacts with HCl(aq) in the reaction: Zn(s) + 2HCI(aq, 4.00 M) → ZnCl,(aq, 0.100 M) + H,(g, 0.100 atm) The immediate amounts of reacting species are: [HCI], = 4.00 M, ZnCl,o = 0.100 M, P, = 0.100 atm. The battery, which utilizes the above redox process possesses the following cell diagram: Zn|Zn*(0.100 M)||H* (4.00 M)|H,(0.100 atm)|Pt. Use the table of standard reduction half-cell potentials in your textbook to calculate: (a) the standard cell potential, Eel, (b) the immediate cell potential, Eell at concentrations and pressures described above, and (c) the immediate Gibbs free energy change for the reaction. Enter your answers in the boxes provided with correct units and sig. figs.: The standard Answer (a): cell potential is ): Ece %3D "cell The immediate Answer (b): %3D cell potential is ): Ecell The immediate Gibbs : AG free energy change is Answer (c): %3D Standard Reduction Potentials at 25°C* Table 18.1 Half-Reaction E(V) F.(g) + 2e - 2F (ag) 0,(g) + 2H*(aq) + 2e O2(g) + H,O Co"(aq) + e – Co2*(aq) H,O2(aq) + 2H*(aq) + 2e 2H,O PbO2(s) + 4H*(aq) + SO (aq) + 2e PBSO,(s) + 2H,O Ce*(aq) + e - Ce"(aq) MnO,(aq) + 8H*(aq) + 5e Mn²*(aq) + 4H,O Au*(aq) + 3e- Au(s) C2(g) + 2e – 2CI (aq) Cr,03 (aq) + 14H*(aq) + 6e¯ → 2Cr**(aq) + 7H,O MnO2(s) + 4H*(aq) + 2e Mn²+(aq) + 2H,O O,(g) + 4H*(aq) + 4e → 2H,O Br,(1) + 2e - 2Br (aq) NO, (aq) + 4H*(aq) + 3e NO(g) + 2H2O 2H3²+(aq) + 2e – Hgž*(aq) Hgž*(aq) + 2e → 2Hg(l) Ag*(aq) + e- Ag(s) Fe*(aq) + e –→ Fe2*(aq) O,(8) + 2H*(aq) + 2e¯ → H,O2(aq) MnO,(aq) + 2H,0 + 3e¯ → MnO2(s) + 40H (aq) L(s) + 2e → 21¯(aq) +2.87 +2.07 +1.82 +1.77 +1.70 +1.61 +1.51 +1.50 +1.36 +1.33 +1.23 +1.23 +1.07 +0.96 +0.92 +0.85 +0.80 +0.77 +0.68 +0.59 +0.53 O,(g) + 2H,0 + 4e¯ → 40H¯(ag) +0.40 Cu*(aq) + 2e → Cu(s) AGCI(s) + e → Ag(s) + Cl¯(aq) So (aq) + 4H*(aq) + 2e SO-(g) + 2H2O Cu*(aq) + e Sn(aq) + 2e 2H*(aq) + 2e Pb*(aq) + 2e Sn2*(aq) + 2e Ni²*(aq) + 2e Co"(aq) + 2e PbSO,(s) + 2e Cd²*(aq) + 2e → Cd(s) Fe*(ag) + 2e → Fe(s) Cr*(aq) + 3e¯ –→ Cr(s) Zn*(aq) + 2e → Zn(s) 2H,0 + 2e → H2(g) + 20H¯(aq) Mn2+(aq) + 2e Mn(s) Al*(ag) + 3e → Al(s) Be*(aq) + 2e Be(s) Mg*(aq) + 2e Mg(s) Na*(ag) + e – Na(s) Ca(aq) + 2e Ca(s) Sr*(aq) + 2e – Sr(s) Ba*(aq) + 2e Ba(s) K*(aq) + e → K(s) Li*(aq) + e +0.34 +0.22 +0.20 → Cu*(aq) Sn²*(aq) +0.15 +0.13 H2(g) 0.00 Pb(s) -0.13 Sn(s) -0.14 Ni(s) -0.25 Co(s) -0.28 Pb(s) + SO (aq) -0.31 -0.40 -0.44 -0.74 -0.76 -0.83 -1.18 -1.66 -1.85 -2.37 -2.71 -2.87 -2.89 -2.90 -2.93 -3.05 Li(s) Increasing strength as oxidizing agent Increasing strength as reducing agent