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20 Electrochemistry Solutions to Exercises (b) Pt Pt e Fe³⁺ (c) 5[Fe²⁺(aq) (d) 0.0592 5 (2.5 (0.001) = 6.510 X 10⁻¹⁷ = 6.5 X 10⁻¹⁷ 5 20.99 (a) V Fe(s) Anode Cathode 2Ag(s) Fe²⁺(aq)+2Ag(s) Fe(s) Ag(s) Ag+ Porous separator (b) e⁻ Zn(s) V H₂(g) + H₂ 0 0 Zn(s) 0 0 Pt Zn²⁺ Porous separator (c) Cu Here, both the oxidized and reduced forms of the cathode solution are in the same phase, so we separate them by a comma, and then indicate an inert electrode. 639