Chemical Thermodynamics

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1 CHAPTER 23. Chemical Thermodynamics (a) H 2 O(l) H 2 O(s) (0 C, 1 bar) Because ice is more ordered structurally than liquid water, ice has less positional disorder than liquid water. Thus, when compared at the same temperature and pressure, the entropy of solid water is less than the entropy of liquid water, and so S sys < 0. (b) Cu(s)(1 bar, 500 C) Cu(s)(25 C, 1 bar) A decrease in temperature at fixed pressure decreases the energy and therefore decreases the thermal disorder of the Cu(s). The decrease in thermal disorder means that the entropy decreases, and thus, S sys < 0. (c) 4 Al(s) + 3O 2 (g) 2Al 2 O 3 (s) (25 C, 1 bar) The reaction described by this equation involves the conversion of seven moles of reactants (three moles of which are gas) into two moles of solid, which corresponds to a large decrease in positional disorder; consequently S sys < Sublimation means the direct conversion of a solid to a gas without passing through the liquid phase. (1) solid gas (sublimation) S sys = S sub For the two-stage process (2) solid liquid gas we have S sys = S fus + S vap. But the initial and final states are the same as in (a) and the entropy is a state function; therefore, S sub = S fus + S vap Further, because S fus > 0 and S vap > 0, S fus + S vap > S vap. Thus, S sub > S vap The order is mainly based upon the number of atoms because the masses of the oxygen and nitrogen atoms are about the same; thus we have NO(g) < N 2 O(g) < NO 2 (g) < N 2 O 4 (g) < N 2 O 5 (g) Note that N 2 O(g) and NO 2 (g) both have three atoms per molecule, but N 2 O(g) has a lower molecular mass than NO 2 (g), and thus we predict that it has the lower entropy of the pair. 75

2 76 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly Because Hrxn > 0 and S rxn < 0, the reaction is not spontaneous under any conditions. Plants use energy from the sun (photosynthesis) to drive the reaction and produce glucose. A nonspontaneous reaction will only form products upon the input from an external source of the energy needed to drive the reaction The concentrations of Ag + (aq) and (aq) before the addition of 5.0 ml of 0.10 M AgNO 3 (aq) are calculated from the K sp expression But [Ag + ] = [ ] and thus K sp = M 2 = [Ag + ] [ ] [Ag + ] = [ ] = M 2 = M The values of [Ag + ] 0 and [ ] 0 immediately after the addition of of 5.0 ml of 0.10 M AgNO 3 (aq) are [Ag + ] 0 = (5.0 ml)(0.10 M) + (100.0 ml)( M) ml [ ] 0 = (100.0 ml)( M) = M ml = M Thus the value of Q c is Q c = [Ag + ] 0 [ ] 0 = ( M)( M) = M 2 The value of G rxn is calculated using Equation G rxn = RT ln Q ( ) c = ( J K 1 mol M 2 )(298 K) ln =+14 kj mol 1 K c M 2 The positive value of G rxn is consistent with the common-ion effect that states that addition of a common ion to a solution decreases the solubility of an ionic solid K = K p /Q p = 1.90 bar/1.00 bar = 1.90 (unitless) The value of G rxn can be calculated from the value of K by using Equation G rxn = RT ln K = ( J K 1 mol 1 )(298 K) ln( ) = J mol 1 = 503 kj mol 1 The reaction as written is spontaneous at 25 C under standard conditions For the reaction described by the equation we have Ag + (aq, M) + (aq, M) Ag(s) Q = Q c Q c = 1 ([Ag + ] 0 /M)([ ] 0 /M) = 1 = 2.5 ( ) 2 University Science Books, All rights reserved.

3 Chapter 23: Chemical Thermodynamics 77 and because K c = 1/K sp, K = K c K c = 1 = Thus, using Equation 23.22, we obtain for G rxn G rxn = RT ln Q ( ) K = ( J K 1 mol 1 11 )(298 K) ln =+9.41 kj mol The positive value of G rxn tells us that no Ag(s) can form under the given conditions of [Ag + ] 0 and [ ] We have that G rxn = G f [CO(g)] + G f [ 2(g)] G f [CO 2(g)] = (1)( kj mol 1 ) + (1)(0 kj mol 1 ) (1)( kj mol 1 ) = 68.7 kj mol 1 = J mol 1 Using Equation 23.19, ln K = G rxn RT = J mol 1 ( J K 1 mol 1 )(298 K) = Start with (from Example 23-10) K = e 27.7 = G rxn = 29.4 kj mol 1 + ( J K 1 mol 1 )(298 K) ln 1 P CO P 2 H 2 = J mol J mol 1 ln 1 P 3 where P = P CO = P H2. We want G rxn < 0, so we want or Therefore, J mol 1 ln 1 P 3 < J mol 1 ln 1 P < J mol 1 = J mol 1 1 P < e = ( ) 1 1/3 or P > = bar Given that H rxn =+55.8 kj mol 1 for the reaction described by the equation 2H 2 O(l) H 3 O + (aq) + OH (aq)

4 78 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly and that K = at 25 C, we use Equation to compute K at 37 C (human body temperature) ( ) [ K ln = J mol 1 ] 310 K 298 K = J K 1 mol 1 (310 K)(298 K) and K = ( )e 0.87 = at 37 C Therefore, [H 3 O + ] = [OH ] = K 1/2 = M and ph = log([h 3 O + ]/M) = 6.81 Because the reaction described by the equation is endothermic, we predict from Le Châtelier s principle that an increase in temperature will shift the reaction to the right, increasing the concentration of H 3 O + (aq) and decreasing the ph. Thus, our results are consistent with Le Châtelier s principle. University Science Books, All rights reserved.

5 CHAPTER 24. Oxidation-Reduction Reactions (a) Applying Rule 7, we assign each oxygen atom an oxidation state of +2. From Rule 2 with x the oxidation state of each chromium atom, we have 2x + (7)( 2) = 2 or x =+6 Thus, the oxidation states of Cr =+6 and O = 2. (b) Applying Rule 4, we assign each fluorine atom an oxidation state of 1. From Rule 2 with x the oxidation state of the hydrogen atom, we have x + (2)( 1) = 1 or x =+1 Thus, the oxidation states are H =+1 and F = 1. (c) Applying Rule 6, we assign each hydrogen atom an oxidation state of +1. From Rule 2 with x the oxidation state of the nitrogen atom, we have x + (4)(+1) =+1 or x = 3 Thus, the oxidation states are H =+1 and N = 3. (d) Applying Rule 2 with x the oxidation state of each iodine atom, we have Thus, the oxidation state is I = 1/3. 3x = 1 or x = 1/ (a) The chlorine atom is more electronegative than the bromine atom, so, according to Rule 8b, we assign both electrons of the covalent bond to the chlorine atom Br A neutral bromine atom has seven valence electrons and has been assigned six in the formula, so oxidation state of Br in Br = 7 6 =+1 A neutral chlorine atom has seven valence electrons and has been assigned eight in the formula, so oxidation state of in Br = 7 8 = 1 79

6 80 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly (b) In this case, the bromine atom is more electronegative than the iodine atom, so we have Br I Thus, we have an oxidation state of +1 for I and 1 for Br. (c) The Lewis formula for I 2 is I The chlorine atom is more electronegative than the iodine atom, so write I A neutral iodine atom has seven valence electrons and has been assigned six in the formula, so oxidation state of I in I 2 = 7 6 =+1 A neutral chlorine atom has seven valence electrons and each has been assigned eight in the formula, so oxidation state of in I 2 = 7 8 = 1 Notice that the sum of the oxidation states of the atoms equals the overall charge, as it must for an ion. (d) The Lewis formula of Br 3 is Br The chlorine atom is more electronegative than the bromine atom, so write Br Thus, the oxidation state of Br is +3 and that of is The oxidation state of each Mn atom in a Mn 2 O 7 molecule is (Rule 7 and Rule 2) given by 2x + (7)( 2) = 0, or x =+7. Therefore, there is a total of = 56 valence electrons in Mn 2 O 7. The Lewis formula is given in the text The oxidation state of iodine changes from 1 inani ion to 1/3 inani 3 ion. Thus, I is the reducing agent and thus the species oxidized. The oxidation state of oxygen changes from 0 in an O 2 molecule to 2 inaoh ion. Thus, O 2 (aq) is the oxidizing agent and the species reduced The oxidation state of copper changes from +1 inacu + ion to both +2 inacu 2+ ion and to 0 in Cu(s). Thus, the half-reaction equations are Cu + (aq) Cu 2+ (aq) + e Cu + (aq) + e Cu(s) (oxidation) (reduction) University Science Books, All rights reserved.

7 Chapter 24: Oxidation-Reduction Reactions 81 In the overall reaction, the Cu + ion is both oxidized (to Cu 2+ ) and reduced (to Cu). Such self oxidation-reduction reactions are called disproportionation reactions The H + (aq) over the reaction arrow indicates that we have an acidic solution. The unbalanced half-reaction equations are Br BrO 3 (oxidation) MnO 4 Mn2+ (reduction) The two half-reactions are balanced with respect to the elements other than H and O. To balance each half reaction with respect to the oxygen atoms, we add three H 2 O molecules to the left side of the oxidation half reaction and four H 2 O molecules to the right side of the reduction half reaction. 3H 2 O + Br BrO 3 (oxidation) MnO 4 Mn2+ + 4H 2 O (reduction) To balance each half reaction with respect to the hydrogen atoms, we add six H + ions to the right side of the oxidation half reaction and eight H + ions to the left side of the reduction half reaction. 3H 2 O + Br BrO 3 + 6H+ (oxidation) 8H + + MnO 4 Mn2+ + 4H 2 O (reduction) Next we balance the half-reactions with respect to charge by adding six electrons to the right side of the oxidation half reaction and five electrons to the left side of the reduction half reaction. 3H 2 O + Br BrO 3 + 6H+ + 6e (oxidation) 8H + + MnO 4 + 5e Mn H 2 O (reduction) Multipling the oxidation half-reaction equation by 5 and the reduction half-reaction equation by 6 yields 15 H 2 O + 5Br 5 BrO H e (oxidation) 48 H MnO e 6Mn H 2 O (reduction) Adding the above two equations yields 15 H 2 O + 5Br + 48 H MnO 4 5 BrO H+ + 6Mn H 2 O Cancelling like terms and indicating phases yield the final balanced equation. 5Br (aq) + 18 H + (aq) + 6 MnO 4 (aq) 5 BrO 3 (aq) + 6Mn2+ (aq) + 9H 2 O(l) The oxidation state of oxygen in H 2 O 2 is 1. When H 2 O 2 acts as a reducing agent, the oxidation state of oxygen increases. The oxidation state of oxygen in O 2 is zero, which is greater than 1. Thus, we have H 2 O 2 O 2

8 82 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly This equation is balanced with respect to all elements except hydrogen. Because we have an acidic aqueous solution, we balance the hydrogen atoms by adding 2 H + to the right side of the equation H 2 O 2 O 2 + 2H + Finally, we add 2 e to the right to balance the charge. The balanced final half-reaction equation is H 2 O 2 O 2 + 2H + + 2e H 2 O 2 (aq) O 2 (g) + 2H + (aq) + 2e The oxidation state of the iron atom changes from +2 in Fe(OH) 2 (s) to+3 in Fe(OH) 3 (s) and thus iron is oxidized. Because iron is oxidized, the other reactant, O 2 (g), must be reduced. Thus we have Fe(OH) 2 Fe(OH) 3 O 2 (oxidation) (reduction) Because the oxidation half-reaction equation involves a metal hydroxide, we skip steps 3 and 4 and balance it directly by adding an OH to the left and and an electron on the right OH + Fe(OH) 2 Fe(OH) 3 + e Because the solution is alkaline, the reduction half reaction is balanced according to steps 3 and 4 by adding 2 H 2 O to the left side and 4 OH to the right side 2H 2 O + O 2 4OH The charge is balanced by adding 4 e to the left side 2H 2 O + O 2 + 4e 4OH (reduction) We can balance the electrons by multiplying the oxidation half-reaction equation by 4 and then adding the result to the reduction half-reaction equation to obtain 4OH + 4 Fe(OH) 2 + 2H 2 O + O 2 4 Fe(OH) 3 + 4OH Cancellation of like terms and designation of phases yield 4 Fe(OH) 2 (s) + 2H 2 O(l) + O 2 (g) 4 Fe(OH) 3 (s) (a) The oxidation and reduction half-reaction equations are CH 3 CH 2 OH CH 3 COOH (oxidation) Cr 2 O 2 7 Cr 3+ (reduction) University Science Books, All rights reserved.

9 Chapter 24: Oxidation-Reduction Reactions 83 The balancing proceeds as follows: (i) elements other than O and H (ii) oxygen CH 3 CH 2 OH CH 3 COOH (oxidation) Cr 2 O 2 7 2Cr 3+ (reduction) H 2 O + CH 3 CH 2 OH CH 3 COOH (oxidation) Cr 2 O 2 7 2Cr H 2 O (reduction) (iii) hydrogen and charge H 2 O + CH 3 CH 2 OH CH 3 COOH + 4H + + 4e (oxidation) 14 H + + Cr 2 O e 2Cr H 2 O (reduction) (iv) multiplying the oxidation half-reaction equation by 3 and the reduction half-reaction equation by 2 (to balance the electrons) and adding the resulting equations yield 3H 2 O + 3CH 3 CH 2 OH + 28 H + + 2Cr 2 O 2 7 3CH 3 COOH + 12 H + + 4Cr H 2 O (v) cancellation of like terms and designating phases yields the final balanced equation 3CH 3 CH 2 OH(aq) + 16 H + (aq) + 2Cr 2 O 2 7 (aq) 3CH 3COOH(aq) + 4Cr 3+ (aq) + 11 H 2 O(l) (b) The number of millimoles of Cr 2 O 2 7 (aq) consumed is given by ( ) millimoles of Cr 2 O 2 10% 7 consumed = (10.0 ml)( M) = mmol 100% The number of millimoles of CH 3 CH 2 OH(aq) oxidized is millimoles of CH 3 CH 2 OH = ( mmol Cr 2 O 2 7 ) 3 mol CH 3CH 2 OH 2 mol Cr 2 O 2 = mmol 7 The number of milligrams of ethanol is mass of CH 3 CH 2 OH = ( mol)(46.07 g mol 1 ) = g = 0.69 mg

10 CHAPTER 25. Electrochemistry A current of 0.50 amperes corresponds to a current flow of 0.50 coulombs per second. The number of electrons that flow through a cross section in one minute is ( )( ) 1 electron 60 s flow of electrons = (0.50 C s 1 ) = electrons min C 1 min Electrons are produced at the cadmium electrode Cd(s) Cd 2+ (aq) + 2e Electrons are consumed at the indium electrode. In 3+ (aq) + 3e In(s) For the description of the electrochemical cell, see the text Oxidation takes place in a basic solution at the left electrode Zn(s) ZnO(s) (oxidation) To balance this equation, we use the following steps: Zn(s) ZnO(s) + H 2 O(l) 2OH (aq) + Zn(s) ZnO(s) + H 2 O(l) 2OH (aq) + Zn(s) ZnO(s) + H 2 O(l) + 2e Reduction takes place in a basic solution at the right electrode. HgO(s) Hg(l) (reduction) To balance this equation, we use the following steps: H 2 O(l) + HgO(s) Hg(l) H 2 O(l) + HgO(s) Hg(l) + 2OH (aq) H 2 O(l) + HgO(s) + 2e Hg(l) + 2OH (aq) 84

11 Chapter 25: Electrochemistry 85 The overall equation for the reaction is the sum of the oxidation half-reaction equation and the reduction half-reaction equation See the text. Zn(s) + HgO(s) ZnO(s) + Hg(l) The cell will continue to discharge until Q = K, that is until chemical equilibrium is attained. When Q = K, ln(q/k ) = 1, and the cell voltage E = 0, from Equation The value of K for the equation that describes the cell reaction Zn(s) + Cu 2+ (aq) Cu(s) + Zn 2+ (aq) is K = (Example 25-5). Because K is so large, the equilibrium lies far to the right and thus we express the equilibrium concentration of Cu 2+ (aq) and Zn 2+ (aq) as Thus [Cu 2+ ] eq = x [equilibrium concentration of unreacted Cu 2+ (aq)] [Zn 2+ ] eq = M (original [Zn 2+ ]) M [from complete conversion of Cu 2+ (aq) to Zn 2+ (aq)] x [concentration of unreacted Cu 2+ (aq)] = M x K = = [Zn2+ ] eq = M x [Cu 2+ ] eq x x = [Cu 2+ ] eq = M [Zn 2+ ] eq = M From Equation and Example 25-6, we have ( ) V E cell = E cell ln Q ν e M x ( ) V = 1.10 V ln [Zn2+ ]/M 2 [Cu 2+ ]/M = 1.10 V ( ) ln = 1.04 V Reduction takes place at the cathode, so E red = V. We use Equation E cell = E red + E ox = V + E ox = V From which we find that E ox = V. Therefore, using Equation 25.16, E red = E ox = V To liberate H 2 (g) from water, V 2+ (aq) must act as a reducing agent. Therefore, the equation for the reaction is 2V 2+ (aq) + 2H + (aq) 2V 3+ (aq) + H 2 (g)

12 86 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly The value of E cell for this cell equation is calculated using the data from Table 25.3: E cell = E red [H+ H 2 ] + E ox [V2+ V 3+ ] = 0 + ( V) = V > 0 Because E cell is positive, V2+ (aq) is capable of liberating H 2 (g) from an acidic aqueous soluton when Q = 1at25.0 C We use Equation ( ) RT E cell = E cell ln Q ν e F where Q = [K+ ] inside /M [K + ] outside /M = 20 and E cell = 0 V because the cell acts as a concentration cell. Thus ( ) ( J mol 1 K 1 )(310 K) E cell = 0V ln 20 = ( J C 1 )ln20 (1)( C mol 1 ) = 0.08 V = 80 mv The magnitude of E cell is 80 mv to one significant figure The two half reactions can be written as H 2 (g) 2H + (aq, unknown M) + 2e and 2H + (1 M) + 2e H 2 (g) The overall equation is 2H + (1 M) 2H + (aq, unknown M) (a concentration cell) The Nernst equation (Equation 25.12) reads E cell = E cell RT ν e F ln Q Using E cell = 0 V for a concentration cell and Equation H3.9 from Appendix H, we have or RT E cell = 0V log Q 2F ( RT [H + (aq, unknown M)]/M = log 2F [H + (1 M)]/M RT = log([h + ]/M) F RT = ph F ( ) F ph = E cell 2.303RT ) 2 University Science Books, All rights reserved.

13 Chapter 25: Electrochemistry The equation for the overall combustion reaction is C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) Each carbon atom in a C 3 H 8 molecule has an oxidation state of 8/3 and the carbon atom in a CO 2 molecule has an oxidation state of +4. Thus, there is a transfer of 20 electrons [(3)( 8 3 ) (3)(4)]. Using the data in Appendix D, G rxn for the stated equation is given by G rxn = 3 G f [CO 2(g)] + 4 G f [H 2O(l)] G f [C 3H 8 (g)] 5 G f [O 2(g)] = (3)( kj mol 1 ) + (4)( kj mol 1 ) ( 23.4 kj mol 1 ) 0kJ mol 1 = kj mol 1 Now we use the equation G rxn = ν efecell to write E cell = G rxn ν e F = J mol 1 (20)( C mol 1 ) = V The equation for the reaction at the anode says that four moles of electrons are required to produce one mole of O 2 (g). Therefore, ( )( ) 2160 C 1 mol O2 moles O 2 (g) produced = = mol C mol 1 4 mol e The volume of O 2 (g) produced at 1.0 bar and 0 C is given by V = nrt = ( mol)( L bar mol 1 K 1 )(273 K) P 1.0 bar = 0.13 L

14 CHAPTER 26. The Chemistry of the Transition Metals For a M(III) ion to be a d 6 ion, the corresponding M(II) ion must be a d 7 ion. The M(II) d 7 ions are Co(II), Rh(II), and Ir(II), and so the M(III) d 6 ions are Co(III), Rh(III), and Ir(III) (a) The [Fe(CN) 6 ] 3 complex anion contains six CN ligands and thus, if we let x equal the oxidation state of the iron ion, we have x + 6( 1) = 3 and x =+3. (b) The [Pt 6 ] 2 complex anion contains six ligands; thus, the oxidation state, x, of the platinum ion is given by x + (6)( 1) = 2 and x = +4. (c) The [Pt(NH 3 ) 3 3 ] + complex cation contains three neutral NH 3 ligands; and three ligands; thus, the oxidation state, x, of the platinum ion is given by x + 3(0) + 3( 1) =+1 and x = The compound [Ag(NH 3 ) 2 ] 3 [Fe(CN) 6 ] contains both a complex cation and a complex anion. The oxidation state of a silver atom in compounds is always +1, and because NH 3 is a neutral ligand, the overall oxidation state of the cation must be +1, or [Ag(NH 3 ) 2 ] +. Therefore, the charge on the anion is 3 and the complex anion is [Fe(CN) 6 ] 3. The oxidation state, x, of the iron ion in the anion is given by x + (6)( 1) = 3, or x =+3. Thus, the salt is named diamminesilver(i) hexacyanoferrate(iii), where we use the name ferrate for the iron in the anion according to Rule 5 and the entry in Table The complex hexaaquanickel(ii) contains a nickel ion in the +2 oxidation state as indicated by the Roman numeral (II) and six (hexa means six) neutral H 2 O ligands; thus, the formula of the complex cation is [Ni(H 2 O) 6 ] 2+. The complex anion diaquatetrabromochromate(iii) contains a chromium ion in the +3 oxidation state, two neutral water ligands, and four bromide ions (Br ). The net charge on the complex anion is 3 + 2(0) + 4( 1) = 1. Thus, the formula of the complex anion is [Cr(H 2 O) 2 Br 4 ]. It requires two 1 anions to balance the +2 cation; therefore, the formula of the compound is [Ni(H 2 O) 6 ][Cr(H 2 O) 2 Br 4 ] (a) The compound K 2 [Fe(EDTA)] contains the complex anion [Fe(EDTA)] 2. The charge on the EDTA ligand is 4 and thus, the oxidation state of the iron ion, x, is given by x + (1)( 4) = 2orx =+2. The name of the compound is potassium ethylenediaminetetraacetatoferrate(ii). (b) The compound Na[Co(C 2 O 4 ) 2 (en)] contains the complex anion [Co(C 2 O 4 ) 2 (en)]. The charge on an oxalate ion is 2 and an ethylenediamine molecule is neutral. Thus, the oxidation state, x, of the cobalt ion is given by x + 2( 2) + 1(0) = 1, or x =+3. The name of 88

15 Chapter 26: The Chemistry of the Transition Metals 89 the compound is sodium ethylenediaminebis(oxalato)cobaltate(iii), where we use the prefix bis because oxalate is a polydentate ligand (Rule 7) The complex anion has six ligands and is octahedral. The two possible geometric isomers are one with the Br and I trans to each other and one with the Br and I cis to each other: Br Co I Br Co I Any other placement of the six ligands gives a structure that can be rotated such that it will superimpose on one of these two structures and thus is not a separate isomer No. The mirror image of each one is superimposable on itself The ground-state d-electron configuration for V(III), a d 2 ion, is t 2 2 e 0 g The d-electron configurations that can give rise to high-spin and low-spin complexes are d 4, d 5, d 6, and d 7 ; Cr(III) is a d 3 ion; Mn(II) is a d 5 ion, and Cu(III) is a d 8 ion. Thus, of the three cases, only Mn(II) can form high-spin and low-spin complexes (a) The complex anion [Fe(CN) 6 ] 3 contains iron in the Fe(III) oxidation state; Fe(III) is a d 5 ion and CN produces a large 0 value. Thus, we predict a low-spin d-electron configuration of t 5 2g e g. 0 The complex anion has one unpaired electron per formula unit and so is paramagnetic. (b) The complex anion [Co(NO 2 ) 6 ] 3 contains cobalt in the Co(III) oxidation state; Co(III) is a d 6 ion and NO 2 produces a large 0 value. Thus, we predict a low-spin d-electron configuration of t 6 2g e g. 0 The complex anion has no unpaired electrons and so is diamagnetic.

CHEM J-14 June 2014

CHEM J-14 June 2014 CHEM1101 2014-J-14 June 2014 An electrochemical cell consists of an Fe 2+ /Fe half cell with unknown [Fe 2+ ] and a Sn 2+ /Sn half-cell with [Sn 2+ ] = 1.10 M. The electromotive force (electrical potential)