Electrchemistry Review: Reductin: the gaining f electrns Oxidatin: the lss f electrns Reducing agent (reductant): species that dnates electrns t reduce anther reagent. Oxidizing agent (xidant): species that accepts electrns t xidize anther species. Oxidatinreductin reactin (redx reactin): a reactin in which electrns are transferred frm ne reactant t anther. 1. Fr example, the reductin f cerium(iv) by irn(ii): Ce 4+ + Fe 2+ Ce 3+ + Fe 3+ a. The reductin halfreactin is given by... Ce 4+ + e Ce 3+ b. The xidatin halfreactin is given by... Fe 2+ e + Fe 3+
Rules fr Balancing OxidatinReductin Reactins A. Write ut halfreactin. B. Balance the halfreactins by adding H +, OH r H 2 O as needed, maintaining electrical neutrality. C. Cmbine the tw halfreactins such that the number f electrns transferred in each cancels ut when cmbined. D. Fr example, cnsider the fllwing reactin f the perxydisulfate in with manganese in: S 2 O 8 2 + Mn 2+ SO 4 2 + MnO 4 1. The reductin step is... S 2 O 8 2 + 2 e + 2 SO 4 2 (Each sulfur atms ges frm +7 t +6 xidatin state.) 2. The xidatin step is... Mn 2+ + 4 H 2 O 5 e + MnO 4 + 8 H + (Manganese(II) lses 5 electrns, ging frm +2 t +7.)
3. In cmbining the tw equatins, the xidatin step must be multiplied by "2," and the reductin step must be multiplied by "5" t cancel ut the electrns: 2x [Mn 2+ + 4 H 2 O 5 e + MnO 4 + 8 H 1+ ] 5x [S 2 O 8 2 + 2 e + 2 SO 4 2 ] Adding these tw equatins tgether... 5 S 2 O 8 2 + 2 Mn 2+ + 8 H 2 O 10 SO 4 2 + 2 MnO 4 + 16 H + 6. Nte that the halfreactins are chargebalanced befre adding them tgether.
OxidatinReductin Reactins in Electrchemical Cells A. It is pssible t separate the halfreactins f an xidatinreductin reactin in an electrchemical cell. B. Cnsider the fllwing reactin: 2 Ag + + Cu (s) 2 Ag (s) + Cu 2+ 1. The reductin halfreactin is given by... Ag + + e Ag (s) 2. The xidatin halfreactin is given by... Cu (s) 2 e + Cu 2+ C. These reactins can be "separated" in a galvanic cell (als called a vltaic cell r battery):
Schematic Representatin f Cells A. The cpper(ii) sulfate/silver nitrate systems described abve wuld be symblized... Cu CuSO 4 (0.02 M) AgNO 3 (0.02 M) Ag 1. Each vertical line (" ") represents a phase bundary r interface where a ptential develps. 2. Each duble vertical line (" ") represents tw phase bundaries (e.g., the salt bridge). 3. The directin f electrn flw is frm left t right: Cu>Ag. B. The equilibrium expressin fr this galvanic cell wuld be... 2 Ag + + Cu (s) 2 Ag (s) + Cu 2+ a. If the battery, initially with 0.0200 M cpper in and 0.0200 M silver in, is allwed t react t equilibrium, the final cncentratins f these ins at equilibrium is determined experimentally t be 0.0300 M cpper in and 2.7 x 10 9 M silver in. Nte that the calculatin abve requires that the cell vltage be zer (i.e., at equilibrium) and that the cncentratins be determined by analysis.
Electric charge/electrmtive frce Electric charge (q) measured in culmbs (C) Magnitude f charge f a single electrn is 1.602 x 10 19 C Mle f electrns has a charge f: (1.602 x 10 19 C) x (6.022 x 10 23 ml 1 ) = 9.649 x 10 4 C q = n F where n = # mles, F = Faraday cnstant Relate (C) t Quantity f Reactins Ex. If 5.585 g f Fe 3+ were reduced in the reactin belw, hw many culmbs f charge must have been transferred frm V 2 + t Fe 3+? Fe 3+ + V 2+ Fe 2+ + V 3+ 4 C 0.1000 ml e (9.649 x 10 ) = 9.649 x 10 3 C ml e
Wrk, Vltage and Free Energy Wrk (J) = E (V) q E = diff in electrical ptential (ptential difference) Ex. Electrical Wrk, p. 285 Free Energy, G Chapter 6 Review******* G = nfe (Relatin between free energy difference and ptential difference) spntaneus reactin: G <0 and E >0 nnspntaneus reactin: G >0 and E <0 Ohm s Law I = R E where I = current, R = resistance (Ω) Pwer (W) P = wrk s = E x q = E s x q s P = E I Fig. 14.2 Electric current diagram I = E R 3.0 V = = 0.030A = 30mA 100Ω P = E I = (3.0V)(0.030 A) = 90mW
Electrchemical Cell revisited Interpreting Electrde Ptentials always write halfreactins as reductins (IUPAC) aa + bb + ne cc + dd E = E lg [ C] a b n [ A] [ B] c [ D] d Ex. Calculate the electrde ptential f a halfcell cntaining 0.100 M KMnO 4 and 0.0500 M MnCl 2 slutin whse ph is 1.000. E MnO4/Mn2+ = 1.51 MnO + 8 H + 5 e Mn 2+ + 4H 4 2 O + n 2+ [Mn ] [MnO ][H E = E MnO 4 / Mn2 lg + 8 4 ] At ph = 1.000, [H 3 O+] = 1.00 x 10 1 M E = 0.0500.51 lg = 1.42 V 1 5 (0.100)(1.00x10 ) 1 8 * Advice fr Finding Relevant HalfReactins, 144, p. 297
as the tendency fr a halfreactin t prceed in the directin f reductin increases, the electrde ptential f the halfcell als increases determine which substance is better xidizing agent: Cu 2+ + 2e Cu (s) Zn 2+ + 2e Zn (s) E =0.521 V E = 0.762 V Calculating cell vltage E cell = E cathde E ande Ex. Calculate the vltage f the fllwing cell: Zn (s) ZnCl 2 (0.120M) Cl 2 (g) (1.15 atm), KCl (0.105 M) Pt right hand cell: Cl 2 (g) + 2e 2 Cl E 1.36 V Cl 2 / Cl = E = E lg n [ Cl p ] Cl2 2 2 (0.105) = 1.36 lg = 1.42 V 2 1.15 Lefthand cell: Zn2+ + 2e Zn (s) E = 2 + Zn / Znl 0.762 V n 1 [ Zn = E lg = 0.762 lg = 0.789 V 2+ ] 2 1 0.120 E cell = E cathde E ande = 1.42 (0.789) = 2.21 V
Assumptin When electrde in cell is nt knwn t be the cathde r righthand electrde.can assume ne electrde is cathde and cell vltage calculated accrdingly. If crrect calculated vltage (+) and the reactin spntaneus If incrrect calculated vltage () and the spntaneus. Cell reactin will be ppsite t the calculated reactin. Ex. Suppse zinc was the cathde in the cell described abve, accrdingly made the righthand electrde: Pt KCl (0.105 M), Cl 2 (g)(1.15 atm) ZnCl 2 (0.120 M) Zn Calculate the vltage and spntaneus reactin f the cell: E zn = 0.789 V, E Pt = 1.42 V Since zinc is the righthand electrde: E cell = E cathde E ande = 0.789 1.42 = 2.21 V Nt spntaneus in the directin written, but rather in the reverse directin T establish reactin directin: Need nly t determine which halfcell has the mre psitive electrde ptential If bth electrde ptentials are the same, n net driving frce (system equilibrium)
Extent f Reactin The Equilibrium Cnstant A x + B red A red + B x The vltage f a galvanic cell that wuld yield this reactin is: E cell = E A E B At equilibrium, the cell vltage is zer and E A = E B lg K eq = n( E A E B ) *Remember: n is the # f electrns fr the tw halfreactins and E A is the standard electrde ptential fr the substance being reduced. Ex. Calculate the equilibrium cnstant fr the reactin: 2Fe 3+ + H 3 AsO 3 + H 2 O 2 Fe 3+ H 3 AsO 4 + 2 H + Halfreactins 2 Fe 3+ + 2e 2 Fe 2+ E = 0.771 V H 3 AsO 4 + 2H + + 2e H 3 AsO 3 + H 2 O E = 0.559 V lg K eq = n( E A E B ) 2(0.771 0.559) lg K eq = = 7. 16 K eq = 10 7.16 = 1.4 x 10 7