Question

7. Calculate the Gibbs energy change from the standard cell potential following redox reactions at 350K

(a) \(\quad A g^{\prime}(a q)+M g(s) \leftrightharpoons A g(s)+M g^{2 *}(a q)\)

(b) \(\mathrm{Cu}^{2 \cdot}(\mathrm{aq})+\mathrm{OH}(\mathrm{aq}) \div \mathrm{Cu}^{+1}(\mathrm{aq})+\mathrm{O}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(1)\)

(c) \(\quad \mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{~s})+\mathrm{Fe}^{+1}(\mathrm{aq}) \leftrightarrows \mathrm{Fe}^{-2}(\mathrm{aq})+\mathrm{Mg}(\mathrm{s})+\mathrm{OH}^{-1}(\mathrm{aq})\)

8. Complete the following table and comment on the spontancity each reaction.

Answer 1

4G -FE - Ans-7 (9) AG = Gibbs free Energy change can be calculated as F = 96500 Cmoit Givent reaction Aytcaq) + mys Agls) + mig at (ay) @__-2884 Anode ng Mg 24 (ay) +20 E = -2:36 Cathode aagtag, 2 Ay É° = Eastholde - Eanode = 0.60 - 62-36) Ecev = 3-168 AG* = -nf = - 2x96500 x 3.46 E = 0.80 V AG = -6098505 126 -609.845 Ans. 7 (6 Given Reactions cut ray} + 01ncaq) Cut Pay) + © Øt Holes Cathode 2 cut2 (aq) + ze 24 E = to 3yy Anode 4or (aq) DOL + 24,0 +46" E= 0.you Eº - Ecathode - Eanode = 0.34 -0.407-0.06¥ 9695-4X96500 X-0.06 = 231605 tace = + 23:160KI

An8-7 (6) Given Reaction Mg(OH)2 (8) + fet cag) petacaq) + ing creon Cathode Anode Macome the Mg + 2017 E - 2-364 2 fet (aq) Gefet? the Fº0.77 . Erell = 3 E Cathode - Eanode - 2-36 – 0.77 = -3.13 ago = 6 - 2x96500x -3.13 Tagº = +604.0 ICJ 7