Question 17.5: For the reaction 3Ag(s) + NO3-(aq) + 4H+(aq) → 3Ag+(aq) + NO...

For the reaction

3Ag(s) + NO_{3}^{-}(aq)  +  4H^{+}(aq)  →  3Ag^{+}(aq)  +  NO(g)  +  2H_{2}O

use Table 17.1 to calculate, at 25°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)
→ 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^{-}

ⓐ ΔG° ⓑ K

ANALYSIS
reaction: (3Ag(s) + NO_{3}^{-}(aq)  +  4H^{+}(aq)  →  3Ag^{+}(aq)  +  NO(g)  +  2H_{2}O)
temperature (25°C)
Information given:
Table 17.1 (standard reduction potentials)
Faraday constant (F)
Information implied
ΔG° Asked for

STRATEGY

1. Assign oxidation numbers.
2. Split the redox reaction into oxidation and reduction half-reactions and find E°.
3. Determine the number of electrons cancelled out when balancing the equation to find n.
4. Substitute into Equation 17.2. Note that ΔG° will be in joules since the Faraday constant is in joules.

ΔG° = -nFE°                                      (17.2)

ANALYSIS
From part (a): E° (0.165 V); n (3 mol) Information given:
K Asked for:

STRATEGY

Substitute into Equation 17.3.

E° = \frac{(0.0257  V)}{n}ln K        (at 25°C)          (17.3)

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N: +5 → +2 reduction

Ag: 0 → +1 oxidation

1. oxidation numbers
NO_{3}^{-}(aq)  +  4H^{+}(aq)  +  3e^{-}  →  NO(g)  +  2H_{2}O             E°_{red} = 0.964 V
Ag(s) + e^{-}  →  Ag^{+}(aq)         E°_{ox} = -0.799 V
E° = 0.964 V + (-0.799 V) = 0.165 V
2. half-reactions
Multiply the oxidation half-reaction by 3 to cancel out the three electrons in the reduction
half-reaction.
Three electrons cancel out: n = 3
3. n
ΔG° = -nFE = (-3 mol)(9.648 × 10^{4}\frac{J}{mol  ·  V})(0.165 V) = -4.78 × 10^{4} J = -47.8 kJ 4. ΔG°

E° = \frac{0.0257 V}{n}ln K; 0.165 V = \frac{0.0257 V}{3}ln K; ln K = 19.3
K = e^{19.3} = 2.4 × 10^{8}
K

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