13_REDOX_Intro - Oxidation-Reduction - REDOX A chemical...

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Unformatted text preview: Oxidation-Reduction - REDOX A chemical reaction where there is a net change in the oxidation number of one or more species. Both an oxidation and a reduction must occur during the reaction. The species that loses electrons during the reaction: Causes a reduction. It is oxidized. Bred = Box + n e- A common approach for listing species that undergo REDOX is by using half-reactions. For 2 Fe3" + Zn°(s) w (reduction) Zn°(s) (oxidation) You’ll find this approach useful for a number of reasons. The species that gains electrons during the reaction: Causes an oxidation. It is reduced. on + n e- = Arecl Fe3+ + e' = Fe2+ Both an oxidation and a reduction must occur in a reaction. The oxidizing agent accepts one or more electrons from the reducing agent. on + Bred : Afed + Box 2 Fe3+ + Zn°(s) = 2 Fe2+ + Zn2+ Tables are available which list half reactions as either oxidations or reductions. Willprovide - Standard E0 values to help predict reactions and equilibria. o Other species that participate in the reaction. * Show the relative ability to gain or loss electrons. ,. w- w; it “w a.er ,. m w; z» “Wm”..er- entials E0 v ljalf reaction — _ i=2 + 2H+ +'e- ="2"HF Ce4+ + e' = Ce3+ (in 1N HCI) 02 + 4H+ + 4e- = 2 H20 1.229 Ag" + e' = Ag 0.800 2H+ + 2e- = H2 0.000 Fe2+ + 2e' = Fe -0.440 Zn2+ + 2e' = Zn -0.763 Al3+ + 3e' = Al -1.66 Li+ + e' = Li -3.04 Having a half-reaction table also make producing a balanced chemical reaction very easy. Determine the balanced equation for the reaction of Fe2+ with Cr2072' in an acidic solution. e “w 1er - ‘ h . A“ j it Half reactlons z» “w 1er - I'IS x V w“ W ' a “' n... W was “‘1 W M a Fro the standard reduction potentials, we can obtain the foilowmg: 0507} + 14 Hr + 66 = 2 06+ + THZO Now all we need to do is make sure that the electrons are the same on both sides. Fe3+ + e- = Fe2* Cr2072' + 14H+ + 6e' = 2 Cr?"r + 7H20 This requires that we multiply the Fe half - We know that one of the half reactions must be . , ‘ reaction by sax so: reversed since an oxidation and a reduction must both occur. . _ 6Fe2++Cr2072' + 14H+ = 6Fe3+ + 2Cr3++7H20 - Based on the question, we know that it is the iron _ half reaction that must be an oxidation. “ «"‘o’a Wm" W "m a v' at. W W“ I" v *1 \ a ‘1 “WWW I REDOX reactions - There are two general ways to conduct an oxidation-reduction reaction Mixing oxidant and reductant together Each half reaction is put in a separate ‘half cell' which is connected electrically. Cu2+ + Znts) ‘1;- Cu(s) + Zn2+ This approach does not allow for any control of our reaction. Permits better control over the system. 4- ’ “S.- Electrons are transferred from one half-cell to the other using an external metal conductor. To complete the circuit, a salt bridge is used. salt bridge * ' . I I Electrochemical cells Salt bridge Allows ion migration in solution but prevents extensive mixing of electrolytes. For our example, we have It can be a simple porous disk or a gel Zinc i0” baing Pmduced- saturated with a non-interfering salt like KCI. This is an oxidation so: The electrode is - the anode - is positive (+). —K+ CI' '— KCI CI' is released K+ is released to Zn side as Zn° is to Cu side as Cu2+ converted to Zn2+ is converted to Cu" .. *“»:,- aw: WI“ .. . »_ 4- ’“»:,- \uw: - For our other half cell, we have copper metal being produced. v The 'spontaneous’ reaction ~ Produces electrical energy Electrolytic o Non-spontaneous reaction ' Required electrical energy to occur This is a reduction so: The electrode is - the cathode _ ' is negatiVe H ' For a reversible cell, galvanic reaction can m occur spontaneously and then be reversed Cu2+ + 2e“ Cu '_ - . electrolytically - rechargeable batteries. a...) x mm. w ' ‘1 In”. A . A measure of how willing a species is to gain or lose electrons. Examples of non-reversible reactions If a gas is produced which escapes. 2 H. + 2 e- _ H2 (9) Potential of a cell acting as a cathode compared to the standard hydrogen electrode. If one or more of the species decomposes. Values also require other standard conditions. ; _ Standard potentials are defined using specific The ultimate reference electrode. _ _. concentrations. H2 is constantly bubbled ' _ l _ All dissolved species are at 1 M into a 1 M HC' 5°!”ti0" . : Slightly soluble species must be at saturation. Pt;r H2 (1atm), 1M H+ ll Pt black '_ I 3 Any gas is constantly introduced at 1 atm plate . E” = 0.000000 V Any metal must be in electrical contact All other standard potentials - Any other solids must also be in contact with are then reported relative to SHE - — ; the conducting electrode. a w- w; 1x am 1er entials The standard potential for Cu2+ + 2e' = Cms) is . ' One thing that we would like to know if the +0.334V. _ _ _' spontaneous direction for a reaction. This mans that: R" _ This requires that we determine the EOE". If azsample of copper metal is placed in a 1 M I Since our Standard potentials (E0) are C“ SOIUt'°n= We ” measure a value 'f 0-334V . . "‘ commonly listed as reductions, we’ll base our if compared to; definitions on that. 2H++2€ = HZQ) ~ E =E -E (1 M) (1 atm) I ' GB" onward reverse E0 — o _ 0 cell _ E fonNard E reverse a m Horn wa‘w 1w xwa x... $11 “pt/"yo; Cell potentials CI '_ You know that both an oxidation and a reduction a m Horn wa‘w 1w xwa x... $11 “pt/"yo; Cell potentials For our copper - zinc cell at standard conditions: a. «9‘ ’ “S.- =‘. «9‘ ’ “S.- must occur. E“ rad- 2+ - = -. _ One of your half reactions must be reversed. cu + 29 cue +0334 V Zn2+ + 2e' = Zn° -0.763 V 2 _ The spontaneous or galvanic direction for a . reaction is the one where Ecall is a positive value -- - so. E 1.097 V cell Galvanic reaction at standard conditions. The half reaction with the largest E value will proceed as a reduction. The ter will be reversed - oxidation — Cu2+ + Zn° = Cu° + Zn2+ $ m «gun “and m at m “1‘1 'F m want; a m a,“ “and m «x m “1‘1 1; 1 “Ii/"‘11; Schematic representations Pt. H2 (1m)! H+ (W) This is the SHE. Pt is used to maintain electrical contact so is listed. The pressure of H2 is given in atmospheres. Schematic representations Rather that attempt to draw out an entire cell, a type of shonhand can be used. :«_ mr’ “i_‘,- r :«_ mr’ “i_‘,- r For our copper - zinc cell, it would be: Zn I Zn2* (1M) II Cu2+ (1M) 1 Cu The anode is always on the left I = boundaries II = salt bridge Other conditions like concentration are listed just after each species. Pt. H, (1m)! HCI (0.01mi! Ag’ (at) [Ag A saturated silver solution (1 .8x10‘3M) based on the KSP of AgCl and the [01']. Different KCI concentrations can be used. 0.1 M - least temperature sensitive Saturated - easier to make and maintain. E = 0.244V Using this electrode will alter our measured values since our reference is now 0.2444V not 0.0000V Calornel electrode (SCE) A much more common 3. reference electrode. It much easier to fiber - work with - no gas. Hg CIZ/KCI - - 2 . Hg 1' Hg20|2(sat), KCI I! g KCI is used to maintain KCI solution _' constant ionic strength. _ “asbestos fiber -_ a m HWY “Hamwmn at m “1‘1 tun w; 1 W’5a9; Concentration dependency of E 1 _H- E0 values are based on standard conditions. a. «9‘ ’ “S.- Since the calomel electrode has a value of 0.244V, all of our measurements will be offset by that amount. The E value will vary if any of the concentrations vary from standard conditions. For standard electrode potentials: This effect can be experimentally determined by measuring E verses a standard (indicating) electrode. E = E0 - 0.244 measured Half reaction EOSHE E°scs Ag+ + e— = Ag 0.800 0.556 an+ + 29- = Zn _0_763 _1 007 Theoretically. the electrode potential can be determined by the Nernst equation. pi" If we assume that concentration is proportional to activity and limit our work to 25°C. the equation becomes: 3 Variations of E _ 0.0592 log [A] are based on '1 [l3]b ratio of the concentrations. _ = standard electrode potential = gas constant - 8.314 Jl°mol = absolute temperature = Faraday’s constant - 96000 coul - - = number of electrons involved ' = activity This also includes a conversion from base e to base 10 logs. I Concentration dependency of E - Exa Determine the potential of a Pt indicating electrode if dipped in a solution containing 0.1 M Sn“+ and 0.01M Sn2+. 1;. aim HWY “Hamwmn at m “jw -\ Another example Determine the potential of a Pt indicating electrode if placed in a solution containing 0.05 M Cr2072'and 1.5 M Or“, if pH = 0.00 (as HCI). '. II Concentration dependency ofE Sn4+ + 29' = Sn2+ E0 = 0.15V Cr2072'+ 14 H“ + 66 = zcr3+ + 7 H20 E =0_15V_ 0.0592 '0 0.01 M 0.1M E0 = 1.33 V =O.18V . A m m 1 ‘W‘M’UHVJ t endency of E t We can use variations in E to determine the = Eo _ 0.0592 log |Cr3*|2 '- _' concentration of species in solution. 5 [Cr2072'][H+]14 ' g - potentiometric methods. I The most straightfonrvard example is when an = 1_33 V - 0-0592 log JEF— :1. electrode is produced by placing a metal in 6 (0-05)“ )14 ' contact with a solution of it’s ion. Ag+ + e. Cu2+ + 2e- Zn2+ + 26 A silver metal indicating electrode is dipped . E = E0 - 0.0592 log into a solution containing Ag”. A value of ' ' 1 [A91 +0.692V is measured verses SHE. _ 0.692 = 0.800 + 0.0592 log [Ag] Determine the concentration of Ag” in this ' ' " Iogmgq : -1 _08 solution. _. 0.0592 : -2 Ag+ + e‘ = A9 E0 = OBOOV [Ag+] 1.0x10 M To determine the galvanic Ecell at standard '_ At non-standard conditions, we don’t know COHdItlonS USIDQ radUCtlon Potemlal51 _ _. which will proceed as a reduction until we _ o o calculate each E value. Ec>ell _ E fontvard ' E reverse Steps in determining the spontaneous direction E0forward - half reaction with the largest or - ' " and E Ofa “IL least negative Eo va|ue_ > Calculate the E for each half reaction _ I. > The half reaction with the largest or least E“reverse - half reaction with the smanest . negative E value will proceed as a reduction. or most negative E° value. ; 3; Calculate Em” = Em“rd _ Ereverse Pb” + 26' = Pb -0.126 V Determine the spontaneous direction and EDeli sn2+ + 2e- = Sn -0.136 V for the following system. Pb / Pb2+ (0.01M) If Sn” (2.5M) I Sn At first glance, it would appear that Pb2+ would be reduced to Pb. However, we're not at Ha" 793050“ E standard conditions. Pb2+ + 28' = Pb -0.126 V Sn2+ + 28' = Sn -0.136V We need to determine the actual E for each half reaction before we know what will happen. Note: The above cell notation may or may not be correct. a m “urn waxrmn xw wit want: Calculation of cell potentials For lead: E = -0.126 - = -.185 V a. «9‘ ’ “S.- Sn2+ + 2e' = Sn -O.124 V - (Pb2+ + 2e' = Pb) -(-0.185 V) 0.0592 1 2 '°9 0.01 Sn” + Pb = Sn + Pb” 0.061 V For tin: E = -0.136 - = -0.124 V 1 log—z5 0.0592 So determining the spontaneous Ecell is really nothing more then doing two sets of . . . . calculations rior to the final calculation. Under our conditions, tIn WIII proceed as a p reduction. We now know that changing concentrations will '- '_ '“P'O change Ece”. E is a measure of the . __ _. Determine the potential of a silver electrode in equilibrium conditions of a REDOX reaction. ' 5 a saturated solution of AgCl. It can be used for determining direction and I 1 Ecell at non-standard conditions. E = 0-800V ' 0-0592 '09 [Ag+] Determining the equilibrium constant for an _. , REDOX reaction ‘ Unfortunately, we don t know [Agt]. Can be modified to include other equilibrium - ' We do know that KSP Age] = [Ag+][c|-] expressions. ._ ‘- . = 1.8)(10'1o u . I I . Mi: Since we're dealing with a saturated AgCI solution, we know that: [A91 = [0"] = KSPW E = 0.800V - 0.0592 log = 0.511 v If [Ag*] 54 [CI'], we could use the expression 1l2 KSP E = 0.800V - 0.0592 log For the general equilibrium: nARED + m Box # n on + m BRED we have two half reactions nm - n x m = lowest common denominator for the balanced reaction. In an earlier example, we determined the galvanic direction for the reaction copper and zinc. Cu2+ + 2e' = Cu° Zn2+ + 2e' = Zn° E +0.334 V 1.097 V cell Galvanic reaction at standard conditions. Cu2+ + Zn° = Cu° + Zn2+ '* no nu ‘ m Equilibriumco stants - We can develop equilibrium constants for REDOX based reactions based on the fact at equilibrium: Ecell = 0 : Efomard ' Ereverse or E0 0.0592| Mic 0.0592 F‘- _ 3- nm nm °9 [Am]n '°9 E0 _E0 = 0.0592 log [AmMBBEgm B A “m [ARED]n[BOX]m \K lo K _ nmlE"B - E09) 9 E0 _ 0.0592 A - species reduced B - species oxidized The K for this reaction would be: _ 2 x 2 x 1.097 '°9KEQ ‘ 0.0592 |Zn2*| = 74 = KEQ 1 .32x10 [012+] Essentially, all copper(|l) will be removed from solution. ’W' ...
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13_REDOX_Intro - Oxidation-Reduction - REDOX A chemical...

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