Question

Using the standard reduction potentials listed, calculate the equilibrium constant for each of the following reactions at 298 K.

E°(V) -0.83 +0.88 +1.78 +0.79 Half-Reaction E°(V) Half-Reaction Ag+ (aq) + - Ag(s) +0.80 2 H20(1) + 2 e — H2(8) + 2 OH+ (aq) AgBr(s) + - Ag(s) + Br" (aq) +0.10 HO2 (aq) + H2O(l) + 2 — 30H(aq) AgCl(s) + Ag(s) + Cl(aq) +0.22 H2O2(aq) + 2 H+ (aq) + 2 —> 2 H20(1) Ag(CN)2 (aq) + — Ag(s) + 2 CN (aq) -0.31 Hg2+ (aq) + 2e — 2 Hg(1) AgzCrO4(s) + 2 — 2 Ag(s) + Cro?-(aq) +0.45 2 Hg2+ (aq) + 2 — Hg2+ (aq) AgI(s) + Ag(s) + (aq) -0.15 Hg2+ (aq) + 2 — Hg(1) Ag(S203)23- (aq) + — Ag(s) + 2 820;} - (aq) +0.01 1(s) + 2e - →21(aq) Al3+ (aq) + 3e —> Al(s) -1.66 2 103 (aq) + 12 H+ (aq) + 10 - 12(s) + 6H2O(1) H3AsO4(aq) + 2 H+ (aq) + 2 H 3AsO3(aq) + H2O(1) +0.56 K+ (aq) + - K(s) Ba²+ (aq) + 2e —Ba(s) -2.90 Lit(aq) + — Li(s) Bio+(aq) + 2 H+ (aq) + 3e Bi(s) + H2O(1) +0.32 Mg2+ (aq) + 2e — Mg(s) Brz(1) + 2 – 2 Br" (aq) +1.07 Mn²+(aq) + 2 – Mn(s) 2 BrO3- (aq) + 12 H+ (aq) + 10 e — Brz(1) + 6H2O(1) MnO2(s) + 4H+ (aq) + 2e — Mn²+ (aq) + 2 H20(1) 2 CO2(g) + 2 H+ (aq) + 2 H2C204(aq) -0.49 MnO4 "(aq) + 8 H(aq) + 5e Mn²+ (aq) + 4H2O(1) Ca2+ (aq) + 2 Ca(s) -2.87 MnO4 (aq) + 2 H2O(1) + 3e — MnO2(s) + 4 OH(aq) Cd2+ (aq) + 2 - Cd(s) -0.40 HNO3(aq) + H+ (aq) + NO(g) + H2O(1) Ce+ (aq) + - Ce3+ (aq) +1.61 N2(g) + 4H2O(l) + 40 - 40H"(aq) + N2H4(aq) Cl2(8) + 2 - 2 Cl(aq) +1.36 N2(g) + 5 H+ (aq) + 4e - N,H5+ (aq) 2 HClO(aq) + 2 H+ (aq) + 2 – C12(8) + 2 H2O(1) +1.63 NO2 (aq) + 4H+ (aq) + 3 N O() + 2 H2O(1) CIO (aq) + H2O(l) + 2 => C (aq) + 2OH(aq) +0.89 Na*(aq) + Na(s) 2 C103 (aq) + 12 H+ (aq) + 10e" — C12(8) + 6 H20(1) +1.47 Ni2+ (aq) + 2 Ni(s) Co2+ (aq) + 2e — Co(s) -0.28 02(g) + 4H+ (aq) + 4e - 2 H2O(1) Co3+ (aq) + — Co2+ (aq) +1.84 02(8) + 2 H2O(l) + 4e — 4OH(aq) Cr3+ (aq) + 3e — Cr(s) -0.74 Oz () + 2 H+ (aq) + 2 H2O2(aq) Cr3+ (aq) + Cr2+ (aq) -0.41 03(8) + 2 H(aq) + 2e 0 2(8) + H2O(1) Cr2O72- (aq) + 14 H+ (aq) + 6e 2 Cr3+ (aq) + 7 H20(1) +1.33 Pb2+ (aq) + 2e - → Pb(s) Cro- (aq) + 4H20(1) + 3e — -0.13 PbO2(s) + HSO4 (aq) + 3H+ (aq) + 2 => Cr(OH)3(s) + 5 OH(aq) PbSO4(s) + 2 H2O(1) +0.92 +0.85 +0.54 +1.20 -2.92 -3.05 -2.37 -1.18 +1.23 +1.51 +0.59 +1.00 -1.16 -0.23 +0.96 +1.52 -2.71 -0.28 +1.23 +0.40 +0.68 +2.07 -0.13 +1.69

A) Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s)

Express your answer using two significant figures.

B) Co(s)+2H+(aq)→Co2+(aq)+H2(g)

Express your answer using two significant figures.

C) 10Br−(aq)+2MnO−4(aq)+16H+(aq)→2Mn2+(aq)+8H2O(l)+5Br2(l)

Express your answer using two significant figure.

Answer 1

(A) Fels) + Ni lap » Fe2+ cap + Nics) Feat cap + aé → Fe 18) E = -0.44 Niet cap + ze → Ni (s) E^ - - 0.28 [-0.28 Ecull = -(-0.44)] = 0.16V At 298 K E all = 0.059 log ko 0.16 = 0.059 log Ke Kez Analog (5.4237) = 265294.8 - 2. 7x15

(B) 1018) + 2H"cap- 604Cap + Hilge 10 2t + é cois) 2H+ (aq) + 2e → Heigo 8 =-0.28v E=0.00 Ecull = o-1-0.280) = 0.28v Еш - о. 054 dos ко = = 9.4915 0.28 X 2 0.059 Ko= 3.1 x 109 a 10 Br cap + 2 Mnoy cap + 16Ht Cuq, 20m2 cups + 8H2O + 5 Br(e) E = 1.070 Br le) + zé - 2 Bv Cup Muay cay + 8H+ cap + ae- Mit + 24h 10 E = 1.230 Full = = 1.23 - 1.07 v 0.16V 0.16 V = 0.059 dog ke lo Ke = 7.0 x 100