(a) Iz(s) + 5 Cu²* (aq) + 6 H;O(I) –→ 2 103"(aq) + 5 Cu(s) + 12 H*(aq) (b) Hg²*(aq) + 21 (aq) → Hg(1) + Iz(s) (c) H2SO3(aq) + 2 Mn(s) + 4 H¯ (aq) · S(s) + 2 Mn²*(aq) + 3 H2O(I)

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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(a) Iz(s) + 5 Cu²* (aq) + 6 H;O(I) –→ 2 103"(aq) + 5 Cu(s) + 12 H*(aq)
(b) Hg²*(aq) + 21 (aq) → Hg(1) + Iz(s)
(c) H2SO3(aq) + 2 Mn(s) + 4 H¯ (aq) ·
S(s) + 2 Mn²*(aq) + 3 H2O(I)
Transcribed Image Text:(a) Iz(s) + 5 Cu²* (aq) + 6 H;O(I) –→ 2 103"(aq) + 5 Cu(s) + 12 H*(aq) (b) Hg²*(aq) + 21 (aq) → Hg(1) + Iz(s) (c) H2SO3(aq) + 2 Mn(s) + 4 H¯ (aq) · S(s) + 2 Mn²*(aq) + 3 H2O(I)
Expert Solution
Step 1
  • I2(s) + 5Cu+2 (aq) + 6H2O → 2IO3-(aq) + 5 Cu (s) + 12 H+

        E cell of the reaction = Ecathode – Eanode   …..(1)

        Gibbs energy is obtained by ∆G = -nFEcell.   ….(2)

     I2(s) → 2IO3-(aq)   ,   oxidation, E = 1.20 V

    5Cu+2 (aq) → 5 Cu (s)    , Reduction E = 0.34 V

         E cell = Ecathode - E anode

                   = 0.34 -1.20 = - 0.86 V, Negative Ecell means positive delta G, hence

       non-spontaneous reaction.

Step 2
  • Hg+2(aq) + 2I- (aq) → Hg(l) + I2 (s)

Hg+2(aq) → Hg (l)   Reduction E = 0.789 V

2I- (aq) → I2 (s), oxidation, E = 0.54 V

           E cell = Ecathode - E anode

                     = 0.789 - 0.54 = 0.249 V, positive Ecell means negative ∆G and hence spontaneous reaction.

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