which of the following pairs of substances could be used to construct a single redox electrode...
Examine the following half-reactions and select the strongest oxidizing agent among the substances. [PEC1412-(aq) + 2e + Pt(s) + 4Cl(aq) Ered = 0.755 V RuO4(s) + 8H(aq) + 8e Ru(s) + 4H2O(1) Ered = 1.038 V Fe042 (aq) + 8H*(aq) + 3e Fe3+ (aq) + 4H2O(1) Ered = 2.07 V -> (Show work for full or partial credit.) RuO4(s) Pt(s) Fe04 (aq) [PtC1412" (aq) Ru(s) Fe3+ag)
Half-ReactionE°(V)Fe^(2+)(aq)+2e^(-)rarrFe(s)-0.44Hg^(2+)(aq)+2e^(-)rarrHg(l)0.86Ag^(+)(aq)+e^(-)rarrAg(s)0.80Cu^(2+)(aq)+2e^(-)rarrCu(s)0.34Zn^(2+)(aq)+2e^(-)rarrZn(s)-0.76What is the value of Δ G° (in kJ ) for the following balanced redox reaction at 25.0℃ ?Hg²⁺(a q)+Fe(s) ⟶ Hg(l)+Fe²⁺(a q)a) -40.5 kJb) -81.1 kJc) -125 kJd) -251 kJ
Table provided below for context Please answer all parts that you can. 1. Which electrochemical cell had the greatest voltage? Identify the anode and the cathode for this pair, the measured cell potential, and the calculated Eºcell- 2. Which electrochemical cell had the smallest voltage? Identify the anode and the cathode for this pair, the measured cell potential, and the calculated Eºcell- 3. If the oxidation and reduction half-reactions are separated in a battery, this means the oxidizing agent is...
Consider an electrochemical cell with a zinc electrode immersed in 1.0 M Zn2and a silver electrode immersed in 1.0 M Agt. Zn2+ + 2e- → Zn E° = -0.76 V Agt + e- Ag E = 0.80 V Calculate Eº and AGº for this cell Calculate AGº for the reaction: 2 Zn (g) + O2 (g) + 2 H20 (1) ► 2 Zn2+ (aq) + 4 OH (aq) Reduction Half-Reaction E° (V) O2(g) + 2 H20 (1)+4 e-®4 OH(aq) +0.403...
HQ18.28 Homework • Unanswered Some electrode combinations that the Phoenix spacecraft designers may have considered for use in their design are given in the table below. Assume the ambient Martian temperature is -67.5*C, 205.7 K. What electrode combination (assume 0.1 M solutions) would provide the 1.317 V required by the spacecraft electronics? Use Appendix 11 for your standard cell potentials. А Rh+ +e- Rh/Ni2+ + 2e + NI O B Rh+ + e + Rh/Cr3+ + 3e - Cr C...
need help for half cell potentials pls calculate step by step (NOTE - Remember that the positive electrode is attached to the red wire and the negative electrode is attached to the black wire.) Electrode Systems Used Anode (oxidation) Negative Cathode (reduction) Positive Measured Potential (V) Positive (Ecu) Copper and silver Cu() - Cu2+ + 2e" Ag+ +1e Ag) 0.432 V Zinc + Silver Zn cs + 2n**+ Zé dat + leº nAg (s) 1.484 V Copper & Zinc Zn(s)...
Which of the following metal(s), if coated onto iron, would prevent the corrosion of iron? Standard Electrode Potentials at 25°C Reduction Half-Reaction Fe2+ (aq) +2 e Zn²+(aq) +2e Pb2+ (aq) +2e A13+ (aq) + 3e E° (V) -0.45 Fe(s) → Zn (s) → Pb (s) + Al(s) 오 오오 | Check all that apply. Zn O Pb Ο ΑΙ
Consider the following table of standard electrode potentials for a series of hypothetical reactions in aqueous solution: Reduction Half-Reaction A+ (aq) + e + A(8) B2+ (aq) + 2e → B(s) C18+ (aq) + e + C2+ (aq) D3+ (aq) + 3e + D(s) E° (V) 1.33 0.87 -0.12 -1.59 Part A Which substance is the strongest oxidizing agent? O D3+ (aq) O A+ (aq) O c3+ (aq) OB2+ (aq) Submit Request Answer Part B Which substance is the weakest...
For the reaction below, find answers for the following 3 questions. In acidic solution, O, and Mn2 ions react spontaneously 0;(&) + Mnaq) + H2O(1) 0.8) + MnO (3) + 2H+ (aq) -0.84 V 1. Calculate in Vand enter to 2 decimal places. 2. Calculate AG (kJ/mole) for the reaction shown above and enter to 1 decimal place 3. Calculate Kat 298 K Enter e notation for large or small numbers.eg, enter 6.7e-34 for 6.7*10% 6.7e34 for 6.7*10% Enter a...
Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.774 M and [Sn2 ] = 0.0190 M. Standard reduction potentials can be found here. Reduction Half-Reaction Standard Potential Ered° (V) F2(g) + 2e– → 2F–(aq) +2.87 O3(g) + 2H3O+(aq) + 2e– → O2(g) + 3H2O(l) +2.076 Co3+(aq) + e– → Co2+(aq) +1.92 H2O2(aq) + 2H3O+(aq) + 2e– → 2H2O(l) +1.776 N2O(g) + 2H3O+(aq) + 2e– → N2(g) + 3H2O(l) +1.766 Ce4+(aq) + e– → Ce3+(aq)...