Question 17.4: Using standard potentials from Table 17.1, decide whether at...

Using standard potentials from Table 17.1, decide whether 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^{-}

ⓐ the reaction

2Fe^{3+}(aq)  +  2I^{-}(aq)  →  2Fe^{2+}(aq)  +  I_{2}(s)

will occur.

ⓑ Fe(s) will be oxidized to Fe^{2+} by treatment with hydrochloric acid.

ⓒ a redox reaction will occur when the following species are mixed in acidic solution: Cl^{-},  Fe^{2+},  Cr^{2+},  I_{2}.

ANALYSIS
equation for the reaction 2Fe^{3+}(aq)  +  2I^{-}(aq)  →  2Fe^{2+}(aq)  +  I_{2}(s) Information given:
Table 17.1 (standard reduction potentials) Information implied:
Will the reaction occur? Asked for:

STRATEGY

1. Assign oxidation numbers.
2. Write oxidation and reduction half-reactions. Include E°_{ox} and E°_{red}.
3. Find E°. The reaction will occur if E° > 0.

ANALYSIS
oxidation half-reaction (Fe(s) → Fe^{2+}(aq)  +  2e^{-}) Information given:
Table 17.1 (standard reduction potentials) Information implied:
Will HCl oxidize Fe? Asked for:

STRATEGY

1. HCl(aq) is made up of two ions, H^{+} and Cl^{-}. Since an oxidizing agent is needed (to oxidize Fe to Fe^{2+}), find either H^{+} or Cl^{-} (or both) in the left column of Table 17.1.
2. Write the possible half-reactions.
3. Write the redox reaction and find E°.

ANALYSIS
ions in acidic solution (Cl^{-},  Fe^{2+},  Cr^{2+},  I_{2}) Information given:
Table 17.1 Information implied:
Will a redox reaction occur when the ions are mixed? Asked for:

STRATEGY

1. Check the left column of Table 17.1 to determine which of the ions are oxidizing agents (i.e., they are reduced). Write the reduction half-reactions of the oxidizing agents.
2. Check the right column of Table 17.1 to determine which of the ions are reducing agents (i.e., they are oxidized). Write the reduction half-reactions of the reducing agents.
3. Write all possible combinations of oxidation and reduction half-reactions. The combination(s) that give positive E° values are possible.
4. Write the redox equation(s) for the reaction(s) that occur.

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Fe: + 3 → +2   reduction

I: -1 → 0           oxidation

1. oxidation numbers
2Fe^{3+}(aq)  +  2e^{-}  →  2Fe^{2+}(aq)      E°_{red}  = +0.769 V
2I^{-}(aq)  →  I_{2}(s)  +  2e^{-}              E°_{ox} = -0.534 V
2. half-reactions
E° = 0.769 V + (-0.534 V) = +0.235 V
E° > 0, the reaction will occur at standard conditions.

Only H^{+} appears in the left column. 1. Oxidizing agent
2H^{+}(aq)  +  2e^{-}  →  H_{2}(g)    E°_{red} = 0.000 V
Fe(s)  →  Fe^{2+}(aq)  +  2e^{-}   E°_{ox} = 0.409 V
2. Half-reactions
Fe(s) + 2H^{+}(aq)  →  Fe^{2+}(aq)  +  H_{2}(g)            E° = 0.409 V
E° > 0, HCl will oxidize Fe at standard conditions (Figure 17.6, page 536).
3. Redox reaction

Fe^{2+}(aq)  +  2e^{-} → Fe(s)          E°_{red} = -0.409 V

I_{2}(s)  +  2e^{-}2I^{-}(aq)          E°_{red} = +0.534 V

1. Oxidizing agents
2Cl^{-}(aq)  →  Cl_{2}(g)  +  2e^{-}                          E°_{ox} = -1.360

Cr^{2+}(aq)  →  Cr^{3+}(aq)  +  e^{-}                E°_{ox} = +0.408

Fe^{2+}(aq)  →  Fe^{3+}(aq)  +  e^{-}            E°_{ox} = -0.769

2. Reducing agents
Fe^{2+}  +  Cl^{-} : E° < 0        I_{2}  +  Cl^{-}: E° < 0

Fe^{2+}  +  Cr^{2+} : E° < 0          I_{2}  +  Cl^{2+}: E° > 0        ✓

Fe^{2+}  +  Fe^{2+} : E° < 0          I_{2}  +  Fe^{2+}: E° < 0

3. Possible combinations
I_{2}(s)  +  2Cr^{2+}(aq)  → 2I^{-}(aq)  +  2Cr^{3+}(aq)              E° = 0.942 V 4. Redox reaction
fig 17.6

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