Question

For all of the following experiments, under standard conditions, which species could be spontaneously produced?

Half-Reaction 8° (V) Half-Reaction 8° (V) 2.87 1.99 1.82 1.78 1.70 1.69 1.68 1.60 1.51 1.50 1.46 1.36 1.33 1.23 F2 + 2e - →2F- Ag2+ + e - Agt CO3+ + e - Co2+ H2O2 + 2H+ + 2e - → 2H2O Ce++ + e - Ce3+ PbO2 + 4H+ + SO42- + 2e - → PbSO4 + 2H2O MnO4+ + 4H+ + 3e - MnO2 + 2H2O 104. + 2H+ + 2e — 103- + H2O MnO4+ + 8H+ + 5e — Mn2+ + 4H2O Au3+ + 3e- -Au PbO2 + 4H+ + 2e →→ Pb2+ + 2H20 Cl2 + 2e - →2C1 Cr20-2- + 14H+ + 6e —→ 2Cr3+ + 7H2O O2 + 4H+ + 4e -→ 2H2O MnO2 + 4H+ + 2e → Mn2+ + 2H2O 103- + 6H+ + 5e - 12 + 3H2O Br2 + 2e - 2Br- VO2+ + 2H+ + - VO2+ + H2O AuCl4- + 3e - Au + 4CI- NO3- + 4H+ + 3e — NO + 2H2O ClO2 +e - CIO 2Hg2+ + 2e - → Hg22+ Agt te - Ag Hg22+ + 2e - → 2Hg Fe3+ + e - Fe2+ O2 + 2H+ + 2e - → H202 MnO4- + - MnO42- 12 + 2e - →21- Cut + e - Cu O2 + 2H2O + 4e - 40H- Cu2+ + 2e - → Cu Hg2Cl2 + 2e - → 2Hg + 2Cl- AgCl + e - Ag + Cl- SO42- + 4H+ + 2e - → H2SO3 + H2O Cu2+ te - Cut 2H+ + 2e → H2 Fe3+ + 3e - Fe Pb2+ + 2e - → Pb Sn2+ + 2e - Sn Ni2+ + 2e - Ni PbSO4 + 2e - → Pb + SO42- Cd2+ + 2e - → Cd Fe2+ + 2e → Fe Cr3+ + e- - Cr2+ Cr3+ + 3e - Cr Zn2+ + 2e - → Zn 2H2O + 2e →→ H2 + 2OH- Mn2+ + 2e - Mn Al3+ + 3e - Al H2 + 2e → 2H- Mg2+ + 2e - → Mg La3+ + 3e - La Na+ te → Na Ca2+ + 2e - Ca Ba2+ + 2e - Ba K+ + - K Li+ + e - Li 1.21 0.40 0.34 0.27 0.22 0.20 0.16 0.00 -0.036 -0.13 -0.14 -0.23 -0.35 -0.40 -0.44 -0.50 -0.73 -0.76 -0.83 - 1.18 -1.66 -2.23 -2.37 -2.37 -2.71 -2.76 -2.90 -2.92 -3.05 1.20 1.09 1.00 0.99 0.96 0.954 0.91 0.80 0.80 0.77 0.68 0.56 0.54 0.52

A lead wire is placed in a solution containing Cu²⁺
yes no  Cu
yes no  PbO₂
yes no  No reaction

Crystals of I₂ are added to a solution of NaCl.
yes no  I^-
yes no  No reaction
yes no  Cl₂

A silver wire is placed in a solution containing Cu²⁺
no yes  Cu
no yes  No reaction
no yes  Ag⁺

Answer 1

According to standard reduction potential :

More the standard reduction potential stronger the oxidizing agent and weaker reducing agent.

Lower the standard reduction potential stronger the reducing agent and weaker oxidizing agent.

a) A lead wire is placed in a solution containing Cu²⁺ :

Pb ----> Pb⁺² + 2e^- E_o = +0.13 V

Cu⁺² +2e^- ----> Cu E₀ = +0.34 V

As Cu⁺² have more standard reduction potential so it would be strong oxidizing agent and undergo reduction (gain of electrons) i.e Cu⁺² +2e^- ----> Cu

As Pb have less standard reduction potential so it would be strong reducing agent and undergo oxidation (loss of electrons) Pb ----> Pb⁺² + 2e^-

b) Crystals of I₂ are added to a solution of NaCl

I₂ + 2e^- ----> 2I^- E_o = 0.54 V

Cl₂ + 2e^- ----> 2Cl^- E_o = 1.36  V

2Cl^- ---->  Cl₂ + 2e^- E_o = -1.36 V

From we can see reducing agent strength for I^- is more than Cl^- and oxidising agent strength for Cl₂ is more than I₂ . So iodine cannot replace chloride ions

I₂ + Cl^- ----> No reaction

Hence the products of these spontaneous reaction will not react.

c) A silver wire is placed in a solution containing Cu²⁺ :

Ag⁺ + e- ----> Ag E₀ = 0.80 V

Ag ---> Ag⁺ + e- Eo = -0.80 V

Cu⁺² + 2e^- ----> Cu Eo = 0.34 V

From we can see reducing agent strength for Cu is more than Ag and oxidising agent strength for Ag⁺ is more than Cu⁺² . So silver cannot replace cupric ions

Cu⁺² + Ag ---> No reaction

Hence the products of these spontaneous reaction will not react.