Question 17.6: GRADED Consider a voltaic cell in which the following reacti...

GRADED

Consider a voltaic cell in which the following reaction occurs:

$O_{2}$(g, 0.98 atm) + $4H^{+}$(aq, pH = 1.24) + $4Br^{-}$(aq, 0.15 M) → $2H_{2}O  +   2Br_{2}(l)$

ⓐ Calculate E for the cell at 25°C.

ⓑ When the voltaic cell is at 35°C, E is measured to be 0.039 V. What is E° at 35°C?

ⓐ

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) → $Mn^{2+}(aq)  +  4H_{2}O$ → $PbSO_{4}(s)  +  2H_{2}O$ → $2H_{2}O$ → $Co^{2+}(aq)$ → 2$F^{-}(aq)$	$Li^{+}(aq)  +  e^{-}$ $K^{+}(aq)  +  e^{-}$ $Ba^{2+}(aq)  +  2e^{-}$ $Ca^{2+}(aq)  +  2e^{-}$ $Na^{+}(aq) +  e^{-}$ $Mg^{2+}(aq)  +  2e^{-}$ $Al^{3+}(aq)  +  3e^{-}$ $Mn^{2+}(aq)  +  2e^{-}$ $Zn^{2+}(aq)  +  2e^{-}$ $Cr^{3+}(aq)  +  3e^{-}$ $Fe^{2+}(aq)  +  2e^{-}$ $Cr^{3+}(aq)  +  e^{-}$ $Cd^{2+}(aq)  +  2e^{-}$ $PbSO_{4}(s)  +  2e^{-}$ $Tl^{+}(aq)  +  e^{-}$ $Co^{2+}(aq)  +  2e^{-}$ $Ni^{2+}(aq)  +  2e^{-}$ $AgI(s)  +  e^{-}$ $Sn^{2+}(aq)  +  2e^{-}$ $Pb^{2+}(aq)  +  2e^{-}$ $2H^{+}(aq)  +  2e^{-}$ $AgBr(s)  +  e^{-}$ $S(s)  +  2H^{+}(aq)  +  2e^{-}$ $Sn^{4+}(aq)  +  2e^{-}$ $SO_{4}^{2-}(aq)  +  4H^{+}(aq) +  2e^{-}$ $Cu^{2+}(aq)  +  e^{-}$ $Cu^{2+}(aq)  +  2e^{-}$ $Cu^{+}(aq)  +  e^{-}$ $I_{2}(s)  +  2e^{-}$ $Fe^{3+}(aq)  +  e^{-}$ $Hg_{2}^{2+}(aq)  +  2e^{-}$ $Ag^{+}(aq)  +  e^{-}$ $2Hg^{2+}(aq)  +  2e^{-}$ $NO_{3}^{-}(aq)  +  4H^{+}(aq) +  3e^{-}$ $AuCl_{4}^{-}(aq)  +  3e^{-}$ $Br_{2}(l)  +  2e^{-}$ $O_{2}(g)  +  4H^{+}(aq)  +  4e^{-}$ $MnO_{2}(s)  +  4H^{+}(aq)  +  2e^{-}$ $Cr_{2}O_{7}^{2-}(aq)  +  14H^{+}(aq)  +  6e^{-}$ $Cl_{2}(g)  +  2e^{-}$ $ClO_{3}^{-}(aq)  +  6H^{+}(aq)  +  5e^{-}$ $Au^{3+}(aq)  +  3e^{-}$ $MnO_{4}^{-}(aq)  +  8H^{+}(aq) +  5e^{-}$ $PbO_{2}(s)  +  SO_{4}^{2-}(aq)  +  4H^{+}(aq)  +  2e^{-}$ $H_{2}O_{2}(aq)  +  2H^{+}(aq)  +  2e^{-}$ $Co^{3+}(aq)  +  e^{-}$ $F_{2}(g)  +  2e^{-}$
	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)		$Fe(OH)_{2}(s)  +  2e^{-}$ $2H_{2}O  +  2e^{-}$ $Fe(OH)_{3}(s)  +  e^{-}$ $S(s)  +  2e^{-}$ $NO_{3}^{-}(aq)  +  2H_{2}O  +  3e^{-}$ $NO_{3}^{-}(aq)  +  H_{2}O +  2e^{-}$ $ClO_{4}^{-}(aq)  +  H_{2}O  +  2e^{-}$ $O_{2}(g)  +  2H_{2}O  +  4e^{-}$ $ClO_{3}^{-}(aq)  +  3H_{2}O  +  6e^{-}$ $ClO^{-}(aq)  +  H_{2}O  +  2e^{-}$

STRATEGY

1. Change pH to $[H^{+}]$ and find Q.
2. Assign oxidation numbers, write oxidation and reduction half-reactions, and cancel electrons to find n.
3. Find E°. ($E°_{red}  +  E°_{ox}$)
4. Substitute into the Nernst equation (Equation 17.4) for T = 25°C

E = E° – $\frac{(0.0257  V)}{n}$ln Q        at 25°C            (17.4)
E = E° – $\frac{0.0257}{n}ln Q$

ⓑ

STRATEGY

Substitute into the Nernst equation for any T.

E = E° – $\frac{RT}{nF}ln Q$