Question

Test Yourself Section.

Controlled-Potential Electrolysis What cathode potential is required to reduce 99.99% of 0.10 M Cu to Cu(s)? Is it pos- sible to remove this Cu without reducing 0.10 M Sn in the same solution? Cu2 +2e Cu(s) EXAMPLE 2+ (17-7) E = 0.339 V (17-8) E= -0.141 V Sn2 2et Sn(s) Solution If 99.99% of Cu were reduced, the concentration of remaining Cu 1.0 X 10 5 M, and the required cathode potential would be would be 0.059 16 log 0.19 V E(cathode) 0.339- 2 1.0 x 10 The cathode potential required to reduce Sn2 is 0.059 16 E(cathode. for reduction of Sn2) -0.141 log 2 0.10, -0.17 V ISn We do not expect reduction of Sn at a cathode potential more positive than -0.17 V. The reduction of 99.99% of Cu without reducing Sn2 2+ appears feasible, TEST YOURSELF Will E(cathode) tion SbO + 2H +3e SbO 0.111 V, so reduction should not occur at 0 19 Y) 0.19 V reduce 0.10 M SbO at pH 2 by the reac- Sb(s) + HO, E 0.208 V? (Answer: E(cathode) for

Answer 1

560+ + 2H+ + Ze z 5b18) + 1420 Feely = ² - Doba logo Foot CHAT Here [1860+] = 610M, PI = 2 - [141] = 10-2-M. JEL = 0-208 – og long lobo( 16212 = 0.1096u Ñ Oul1V (Answer Since Ecathode for sbot koolqv so reduction will not take place.

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