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(a) Iz(s) + 5 Cu²* (aq) + 6 H2O(1) (b) Hg²*(aq) + 21 (aq) (c) H2SO3(aq) + 2 Mn(s) + 4 H*(aq) → S(s) + 2 Mn²* (aq) + 3 H,O(I) 2 10;"(aq) + 5 Cu(s) + 12 H*(aq) - Hg(1) + I2(s) - -

(a) Iz(s) + 5 Cu²* (aq) + 6 H2O(1) (b) Hg²*(aq) + 21 (aq) (c) H2SO3(aq) + 2 Mn(s) + 4 H*(aq) → S(s) + 2 Mn²* (aq) + 3 H,O(I) 2 10;"(aq) + 5 Cu(s) + 12 H*(aq) - Hg(1) + I2(s) - -

Chemistry for Engineering Students
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter4: Stoichiometry
Section: Chapter Questions
Problem 4.98PAE
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(a) Iz(s) + 5 Cu²* (aq) + 6 H2O(1) (b) Hg²*(aq) + 21 (aq) (c) H2SO3(aq) + 2 Mn(s) + 4 H*(aq) → S(s) + 2 Mn²* (aq) + 3 H,O(I) 2 10;"(aq) + 5 Cu(s) + 12 H*(aq) - Hg(1) + I2(s) - -
Transcribed Image Text:(a) Iz(s) + 5 Cu²* (aq) + 6 H2O(1) (b) Hg²*(aq) + 21 (aq) (c) H2SO3(aq) + 2 Mn(s) + 4 H*(aq) → S(s) + 2 Mn²* (aq) + 3 H,O(I) 2 10;"(aq) + 5 Cu(s) + 12 H*(aq) - Hg(1) + I2(s) - -
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, Eₒ = -1.20 V

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

         E cell = Ecathode – E anode

                   = 0.34 – (- 1.20) = 1.54 V, which is the positive potential , hence ∆G becomes negative and reaction will be spontaneous.

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

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

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

           E cell = Ecathode – E anode

                     = 0.789 – (- 0.54) V = 1.329 V , positive E cell means negative ∆G and spontaneous reaction.

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