Relating ΔG° and [latex]E^\circ _{cell}[/latex] Use the tabulated electrode potentials to calculate ∆G° for the reaction: [latex]I_2(s) + 2 Br^-(aq) \longrightarrow 2 I^-(aq) + Br_2(l)[/latex] Is the reaction spontaneous?

Question 19.6: Relating ΔG° and E°cell Use the tabulated electrode potentia...

Relating ΔG° and $E^\circ _{cell}$

Use the tabulated electrode potentials to calculate ∆G° for the reaction:

$I_2(s) + 2 Br^-(aq) \longrightarrow 2 I^-(aq) + Br_2(l)$

Is the reaction spontaneous?

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

SORT You are given a redox reaction and asked to find ∆G°.

GIVEN:

I_2(s) + 2 Br^-(aq) \longrightarrow 2 I^-(aq) + Br_2(l )

FIND: ∆G°

STRATEGIZE Refer to the values of electrode potentials in Table 19.1 to calculate

E_{cell}^\circ

Then use Equation 19.3 to calculate ∆G° from $E_{cell}^\circ$ .

$∆G^\circ= −nFE_{cell}^\circ$ [19.3]

CONCEPTUAL PLAN

E_{an}^\circ ,E_{cat}^\circ \longrightarrow E_{cell}^\circ

$E_{cell}^\circ \underset{∆G^\circ= −nFE_{cell}^\circ}{\longrightarrow} \Delta G\circ$

SOLVE Separate the reaction into oxidation and reduction half-reactions and find the standard electrode potentials for each. Determine

E_{cell}^\circ

by subtracting

E_{an}

from

E_{cat}

SOLUTION

Oxidation (Anode): $2 Br^-(aq) \longrightarrow Br_2(l ) + \cancel{2 e^-} E^\circ = 1.09 V$

Reduction (Cathode): $I_2(s) + \cancel{2 e^-} \longrightarrow 2 I^-(aq) E^\circ = 0.54 V \\ \overline{I_2(s) + 2 Br^-(aq) \longrightarrow 2 I^-(aq) + Br_2(l) E_{cell}^\circ=E_{cat}^\circ -E_{an}^\circ = -0.55 V}$

Calculate ∆G° from

E_{cell}^\circ

The value of n (the number of moles of electrons) corresponds to the number of electrons that are canceled in the half-reactions. Remember that 1 V = 1 J/C.

∆G^\circ= −nFE_{cell}^\circ

= -2  \cancel{mol}  e^- (\frac{ 96,485  \cancel{C}}{\cancel{mol}  e^-}) ( -0.55\frac{J}{\cancel{C}})

= +1.1 × $10^5$ J

Since ∆G° is positive, the reaction is not spontaneous under standard conditions.

CHECK The answer is in the correct units (joules) and seems reasonable in magnitude (≈110 kJ). You have seen (in Chapter 18) that values of ∆G° typically range from plus or minus tens to hundreds of kilojoules. The sign is positive, as expected for a reaction in which

E_{cell}^\circ

is negative.

Table 19.1

Standard Electrode Potentials at 25 °C
Reduction Half-Reaction			E°(V)
	$F_2(g) + 2 e^-$	$\longrightarrow 2 F^-(aq)$	2.87
	$H_2O_2(aq) + 2 H^+(aq) + 2 e^-$	$\longrightarrow 2 H_2O(l)$	1.78
	$PbO_2(s) + 4 H^+(aq) + SO_4{}^{2-}(aq) + 2 e^-$	$\longrightarrow PbSO_4(s) + 2 H_2O(l)$	1.69
	$MnO_4^-(aq) + 4 H^+(aq) + 3 e^-$	$\longrightarrow MnO_2(s) + 2 H_2O(l)$	1.68
	$MnO_4{}^-(aq) + 8 H^+(aq) + 5 e^-$	$\longrightarrow Mn^{2 +}(aq) + 4 H_2O(l)$	1.51
	$Au^{3 +}(aq) + 3 e^-$	$\longrightarrow Au(s)$	1.50
	$PbO_2(s) + 4 H^+(aq) + 2 e^-$	$\longrightarrow Pb^{2 +}(aq) + 2 H_2O(l)$	1.46
	$Cl_2(g) + 2 e^-$	$\longrightarrow 2 Cl^-(aq)$	1.36
	$Cr_2O_7{}^{2 -}(aq) + 14 H^+(aq) + 6 e^-$	$\longrightarrow 2 Cr^{3 +}(aq) + 7 H_2O(l)$	1.33
	$O_2(g) + 4 H^+(aq) + 4 e^-$	$\longrightarrow 2 H_2O(l)$	1.23
	$MnO_2(s) + 4 H^+(aq) + 2 e^-$	$\longrightarrow Mn^{2 +}(aq) + 2 H_2O(l)$	1.21
	$IO_3^-(aq) + 6 H^+(aq) + 5 e^-$	$\longrightarrow \frac{1}{2} I_2(aq) + 3 H_2O(l)$	1.20
	$Br_2(l) + 2 e^-$	$\longrightarrow 2 Br^-(aq)$	1.09
	$VO_2^{+}(aq) + 2 H^+(aq) + e^-$	$\longrightarrow VO^{2 +}(aq) + H_2O(l)$	1.00
	$NO_3{}^{-}(aq) + 4 H^+(aq) + 3 e^-$	$\longrightarrow NO(g) + 2 H_2O(l)$	0.96
	$ClO_2(g) + e^-$	$\longrightarrow ClO_2{}^-(aq)$	0.95
	$Ag^+(aq) + e^-$	$\longrightarrow Ag(s)$	0.80
	$Fe^{3 +}(aq) + e^-$	$\longrightarrow Fe^{2 +}(aq)$	0.77
	$O_2(g) + 2 H^+(aq) + 2 e^-$	$\longrightarrow H_2O_2(aq)$	0.70
	$MnO_4{}^-(aq) + e^-$	$\longrightarrow MnO_4{}^{2-}(aq)$	0.56
	$I_2(s) + 2 e^-$	$\longrightarrow 2 I^-(aq)$	0.54
	$Cu^+(aq) + e^-$	$\longrightarrow Cu(s)$	0.52
	$O_2(g) + 2 H_2O(l) + 4 e^-$	$\longrightarrow 4 OH^-(aq)$	0.40
	$Cu^{2+}(aq) + 2 e^-$	$\longrightarrow Cu(s)$	0.34
	$SO_4{}^{2 -}(aq) + 4 H^+(aq) + 2 e^-$	$\longrightarrow H_2SO_3(aq) + H_2O(l)$	0.20
	$Cu^{2+}(aq) + e^-$	$\longrightarrow Cu^+(aq)$	0.16
	$Sn^{4+}(aq) + 2 e^-$	$\longrightarrow Sn^{2 +}(aq)$	0.15
	$2 H^+(aq) + 2 e^-$	$\longrightarrow H_2(g)$	0
	$Fe^{3+}(aq) + 3 e^-$	$\longrightarrow Fe(s)$	–0.036
	$Pb^{2+}(aq) + 2 e^-$	$\longrightarrow Pb(s)$	-0.13
	$Sn^{2+}(aq) + 2 e^-$	$\longrightarrow Sn(s)$	-0.14
	$Ni^{2+}(aq) + 2 e^-$	$\longrightarrow Ni(s)$	-0.23
	$Cd^{2+}(aq) + 2 e^-$	$\longrightarrow Cd(s)$	-0.40
	$Fe^{2+}(aq) + 2 e^-$	$\longrightarrow Fe(s)$	-0.45
	$Cr^{3+}(aq) + e^-$	$\longrightarrow Cr^{2+}(aq)$	-0.50
	$Cr^{3+}(aq) + 3 e^-$	$\longrightarrow Cr(s)$	-0.73
	$Zn^{2+}(aq) + 2 e^-$	$\longrightarrow Zn(s)$	-0.76
	$2 H_2O(l) + 2 e^-$	$\longrightarrow H_2(g) + 2 OH^-(aq)$	-0.83
	$Mn^{2+}(aq) + 2 e^-$	$\longrightarrow Mn(s)$	-1.18
	$Al^{3+}(aq) + 3 e^-$	$\longrightarrow Al(s)$	-1.66
	$Mg^{2+}(aq) + 2 e^-$	$\longrightarrow Mg(s)$	-2.37
	$Na^+(aq) + e^-$	$\longrightarrow Na(s)$	-2.71
	$Ca^{2+}(aq) + 2 e^-$	$\longrightarrow Ca(s)$	-2.76
	$Ba^{2+}(aq) + 2 e^-$	$\longrightarrow Ba(s)$	-2.90
	$K^+(aq) + e^-$	$\longrightarrow K(s)$	-2.92
	$Li^+(aq) + e^-$	$\longrightarrow Li(s)$	-3.04