Half-ReactionE"(V)Half-ReactionE(V)Ag *(aq) + eAgBr(s) + eAgCI(s) + eAg(CN), (aq) + e¯Ag,CrO,(s) + 2 eAgI(s) + e -Ag(s)Ag(s) + Br (ag)Ag(s) + CI"(ag)2 H,0(1) + 2 eHO, (ag) + H,0(1) + 2 e - 3 OH (ag)H,O,(aq) + 2 H*(aq) + 2 e -Hg,2* (ag) + 2 e-2 Hg* (aq) + 2 eHg**(aq) + 2 eI(s) + 2 e+0.799H2(g) + 2 OH (aq)--0.83+0.095+0.88>+0.2222 H;O(1)+1.776Ag(s) + 2 CN (aq)-0.312 Hg(1)Hg,2+ (aq)Hg()21 (aq)->+0.7892 Ag(s) + Cro, (ag)-+0.446w+0.920Ag(s) + 1(aq)Ag(S,O3)2 (aq) + e →Al* (ag) vt 3 e Al(s)H3ASO4(aq) + 2 H* (ag) + 2 e --0.151+0.854Ag(s) + 2 S,0,"(aq)+0.01+0.536-1.662 10; (aq) + 12 H*(aq) + 10 e+1.195+0.55912(s) + 6 H2O()H3ASO;(ag) + H,0(1)K*(aq) + eLi*(ag) + eMg?*(ag) + 2 e-Mn2+ (aq) + 2 e“K(s)Ba2+ (aq) + 2 eBio* (aq) + 2 H* (aq) + 3 e-Br2(1) + 2 e - 2 Br (aq)2 Bro; (ag) + 12 H*(aq) + 10 e --2.925Ba(s)-2.90Bi(s) + H2O(1)Li(s)-3.05+0.32+1.065Mg(s)-2.37+1.52» Mn(s)-1.18MnO;(s) + 4 H*(aq) + 2 e-Brz(1) + 6 H2O(1)H2C,O4(ag)+1.23>Mn*(aq) + 2 H;0()2 CO:(g) + 2 H* (aq) + 2e →Ca2+ (aq) + 2 eCd2+(aq) + 2 eCe*+(aq) + e →Cl(g) + 2 e" - 2 CI (ag)2 HCIO(ag) + 2 н*(аq) + 2 е-0.49MnO4 (aq) + 8 H* (aq) + 5 e --→Ca(s)Cd(s)Ce* (aq)--2.87+1.51Mn?*(ag) + 4 H;O()-0.403Mno, (ag) + 2 H,O(1) + 3 e+0.59+1.61MnO2(s) + 4 OH"(aq)+1.359HNO,(aq) + H* (aq) + e"N2(g) + 4 H20(!) + 4eNO(g) + H2O(1)4 OH (aq) + N,H4(aq)N,H3*(aq)+1.63+1.00Cl2(g) + 2 H2O(I)-1.16CIO (aq) + H,O(1) + 2 eN2(g) + 5 H* (ag) + 4 eNO, (aq) + 4 H* (aq) + 3 e --→ NO(g) + 2 H,O()Na* (ag) + e – Na(s)Ni?* (aq) + 2 eOz(g) + 4 H*(aq) + 4eOz(g) + 2 H;O(I) + 4 e 4 OH (aq)O,(g) + 2 H* (ag) + 2eO3lg) + 2 H*(aq) + 2 ePb* (aq) + 2 e Pb(s)PbO2(s) + HSO, (aq) + 3 H*(aq) + 2 e+0.89-0.23CI (ag) + 2 OH (aq)+0.962 ClO, (ag) + 12 H* (aq) + 10 e+1.47-2.71Cl2(g) + 6 H20(1)- Ni(s)-0.28Co2+(aq) + 2 e Co(s)Co+ (aq) + e Co2+(aq)Cr*(aq) + 3 e"Cr*(ag) + eCro,"(aq) + 14 H*(aq) + 6 e --0.2772 H,O(1)+1.23+1.842+0.40H;O2(aq)O,(g) + H,0()- Cr(s)-0.74+0.68Cr* (aq)+2.07-0.41+1.33-0.1262 Cr* (aq) + 7 H,0(1)+1.685Cro,"(aq) + 4 H;O() + 3e -PbSO,(s) + 2 H0(1)Pb(s) + HSO, (aq)-0.13Cr(OH)3(s) + 5 OH (aq)PbSO,(s) + H*(aq) + 2 ePtCl (aq) + 2 eS(s) + 2 H*(aq) + 2 eH,SO3(ag) + 4 H*(aq) + 4 e - S(s) + 3 H2O(1)HSO, (aq) + 3 H*(aq) + 2 e-0.356Cu?*(ag) + 2 e Cu(s)Cu2* (ag) + e - Cu* (aq)Cu*(aq) + eCul(s) + e+0.337Pt(s) + 4 CI (aq)+0.73+0.153H,S(g)+0.141- Cu(s)- Cu(s) + I (aq)2 F(aq)+0.521+0.45-0.185+0.17F2(g) + 2 eFe*(ag) + 2 eFe*(ag) + e →Fe(CN), (aq) + e2 H* (ag) + 2 e →-+2.87H2SO3(aq) + H,0(1)Sn²*(aq) + 2 eSn** (aq) + 2 e-→VO," (ag) + 2 H*(aq) + eZn2+(aq) + 2 eFe(s)-0.440Sn(s)--0.136Fe2* (aq)Fe(CN), (aq)+0.771Sn²* (aq)+0.154+0.36vo* (ag) + H,O(1) +1.00H;(g)0.000Zn(s)-0.763 4. *For the unbalanced reaction Fe3+(aq) +I-(aq) → Fe²*(aq) + I2(s)а.Balance the reaction and label the oxidation and reduction half reactions.b. Identify the anode and cathodec. What is the oxidizing agent? The reducing agent?d. Using the table of standard reduction potentials, calculate E° for this cell.

Basic terms: Anode: It is an electrode at which the oxidation half-reaction occurs. Cathode: It is…

4. *For the unbalanced reaction Fe3+(aq) +I-(aq) → Fe²*(aq) + I2(s) Balance the reaction and label the oxidation and reduction half reactions. b. Identify the anode and cathode c. What is the oxidizing agent? The reducing agent? d. Using the table of standard reduction potentials, calculate E° for this cell.

4. For the unbalanced reaction Fe3+(aq) +I-(aq) → Fe²(aq) + I2(s) Balance the reaction and label the oxidation and reduction half reactions. b. Identify the anode and cathode c. What is the oxidizing agent? The reducing agent? d. Using the table of standard reduction potentials, calculate E° for this cell.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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A chemist designs a galvanic cell that uses these two half-reactions: half-reaction + 21O3(aq)+ 12 H¹ (aq)+10e¯ Cl₂(g) +2e 1₂(s) + 6H₂O(1) → 2 C1 (aq) standard reduction potential E=+1.195 V 'red E 70 = +1.359 V red