Question 21.3: Using Table 17.1, decide whether hydrogen peroxide will reac...

Using Table 17.1, decide whether hydrogen peroxide will react with (oxidize or reduce) the following ions in acid solution at standard concentrations.

Table 17.1 Standard Potentials in Water Solution at 25°C

Lithium is the strongest reducing agent.               O = strongest oxidizing agent; R = strongest reducing agent.                       Fluorine is the strongest oxidizing agent.   Lithium and fluorine are very dangerous materials to work with.	AcidicSolution,[H^{+}] = 1 M
	E°_{red}(V)
	-3.040 -2.936 -2.906 -2.869 -2.714 -2.357 -1.680 -1.182 -0.762 -0.744 -0.409 -0.408 -0.402 -0.356 -0.336 -0.282 -0.236 -0.152 -0.141 -0.127 0.000 0.073 0.144 0.154 0.155 0.161 0.339 0.518 0.534 0.769 0.796 0.799 0.908 0.964 1.001 1.077 1.229 1.229 1.330 1.360 1.458 1.498 1.512 1.687 1.763 1.953 2.889		→Li(s) → K(s) → Ba(s) → Ca(s) →Na(s) → Mg(s) → Al(s) → Mn(s) → Zn(s) → Cr(s) → Fe(s) → Cr^{2+}(aq) → Cd(s) → Pb(s) + SO_{4}^{2-}(aq) → Tl(s) → Co(s) →Ni(s) → Ag(s) + I^{-}(aq) → Sn(s) → Pb(s) →H_{2}(g) → Ag(s) + Br^{-}(aq) →H_{2}S(aq) → Sn^{2+}(aq) → SO_{2}(g)  +  2H_{2}O → Cu^{+}(aq) → Cu(s) → Cu(s) → 2I^{-}(aq) → Fe^{2+}(aq) → 2Hg(l) → Ag(s) →Hg_{2}^{2+}(aq) →NO(g) + 2H_{2}O → Au(s)  +  4Cl^{-}(aq) → 2Br^{-}(aq) → 2H_{2}O → Mn^{2+}(aq)  +  2H_{2}O → 2Cr^{3+}(aq)  +  7H_{2}O → 2Cl^{-}(aq) →\frac{1}{2} Cl_{2}(g)  +  3H_{2}O → Au(s) → Mn^{2+}(aq)  +  4H_{2}O → PbSO_{4}(s)  +  2H_{2}O → 2H_{2}O → Co^{2+}(aq) → 2F^{-}(aq)	Li^{+}(aq)  +  e^{-} K^{+}(aq)  +  e^{-} Ba^{2+}(aq)  +  2e^{-} Ca^{2+}(aq)  +  2e^{-} Na^{+}(aq) +  e^{-} Mg^{2+}(aq)  +  2e^{-} Al^{3+}(aq)  +  3e^{-} Mn^{2+}(aq)  +  2e^{-} Zn^{2+}(aq)  +  2e^{-} Cr^{3+}(aq)  +  3e^{-} Fe^{2+}(aq)  +  2e^{-} Cr^{3+}(aq)  +  e^{-} Cd^{2+}(aq)  +  2e^{-} PbSO_{4}(s)  +  2e^{-} Tl^{+}(aq)  +  e^{-} Co^{2+}(aq)  +  2e^{-} Ni^{2+}(aq)  +  2e^{-} AgI(s)  +  e^{-} Sn^{2+}(aq)  +  2e^{-} Pb^{2+}(aq)  +  2e^{-} 2H^{+}(aq)  +  2e^{-} AgBr(s)  +  e^{-} S(s)  +  2H^{+}(aq)  +  2e^{-} Sn^{4+}(aq)  +  2e^{-} SO_{4}^{2-}(aq)  +  4H^{+}(aq) +  2e^{-} Cu^{2+}(aq)  +  e^{-} Cu^{2+}(aq)  +  2e^{-} Cu^{+}(aq)  +  e^{-} I_{2}(s)  +  2e^{-} Fe^{3+}(aq)  +  e^{-} Hg_{2}^{2+}(aq)  +  2e^{-} Ag^{+}(aq)  +  e^{-} 2Hg^{2+}(aq)  +  2e^{-} NO_{3}^{-}(aq)  +  4H^{+}(aq) +  3e^{-} AuCl_{4}^{-}(aq)  +  3e^{-} Br_{2}(l)  +  2e^{-} O_{2}(g)  +  4H^{+}(aq)  +  4e^{-} MnO_{2}(s)  +  4H^{+}(aq)  +  2e^{-} Cr_{2}O_{7}^{2-}(aq)  +  14H^{+}(aq)  +  6e^{-} Cl_{2}(g)  +  2e^{-} ClO_{3}^{-}(aq)  +  6H^{+}(aq)  +  5e^{-} Au^{3+}(aq)  +  3e^{-} MnO_{4}^{-}(aq)  +  8H^{+}(aq) +  5e^{-} PbO_{2}(s)  +  SO_{4}^{2-}(aq)  +  4H^{+}(aq)  +  2e^{-} H_{2}O_{2}(aq)  +  2H^{+}(aq)  +  2e^{-} Co^{3+}(aq)  +  e^{-} F_{2}(g)  +  2e^{-}
	Basic Solution, [OH^{-}] = 1 M
	E°_{red}(V)
	-0.891 -0.828 -0.547 -0.445 -0.140 0.004 0.398 0.401 0.614 0.890	→ Fe(s) + 2 OH^{-}(aq) →H_{2}(g) + 2 OH^{-}(aq) → Fe(OH)_{2}(s) + OH^{-}(aq) → S^{2-}(aq) →NO(g) + 4 OH^{-}(aq) →NO_{2}^{-}(aq) + 2 OH^{-}(aq) → ClO_{3}^{-}(aq) + 2 OH^{-}(aq) → 4 OH^{-}(aq) → Cl^{-}(aq) + 6 OH^{-}(aq) → Cl^{-}(aq) + 2 OH^{-}(aq)		Fe(OH)_{2}(s)  +  2e^{-} 2H_{2}O  +  2e^{-} Fe(OH)_{3}(s)  +  e^{-} S(s)  +  2e^{-} NO_{3}^{-}(aq)  +  2H_{2}O  +  3e^{-} NO_{3}^{-}(aq)  +  H_{2}O +  2e^{-} ClO_{4}^{-}(aq)  +  H_{2}O  +  2e^{-} O_{2}(g)  +  2H_{2}O  +  4e^{-} ClO_{3}^{-}(aq)  +  3H_{2}O  +  6e^{-} ClO^{-}(aq)  +  H_{2}O  +  2e^{-}

(a) I^{-}          (b) Sn^{2+}             (c) Co^{2+}

STRATEGY

1. Since H_{2}O_{2} disproportionates, it can either be reduced (and oxidize the ions) or be oxidized (and reduce the ions).
2. For a redox reaction to occur where H_{2}O_{2} is reduced,
E°_{red}  H_{2}O_{2}(1.763 V) + E°_{ox}  ion > 0
3. For a redox reaction to occur where H_{2}O_{2} is oxidized,
E°_{ox}  H_{2}O_{2}(-0.695 V) + E°_{red} ion > 0