Given the standard electrode reduction potentials, which reaction will occur spontaneously? Standard Reduction E° (V) Potentials Cu2+(aq) + e- → Cu+(aq) +0.15 V2+(aq) + 2e- V(s) -1.18 Mg2+(aq) + 2e- → Mg(s) -2.37 A) V2+ + 2Cu* → V + 2Cu2+ B) Mg2+ + V > V²* + Mg C) Mg2+ + 2Cu* → 2Cu2* + Mg D) Mg2+ + V2*→ Mg + V

Standard Electrode Potentials at 25 °C Reduction Half-Reaction O Cu (aq) O H,Ozlaq) O Br O 10₂ (aq) Fyl) +20 H₂O₂(aq) + 2 H(aq) +2e" PbOg(s) +4 H(aq) +50, (aq) + 2 e Mn0 (aq) +4 H(aq) + 3 e MnO, (aq) +8H(aq) + 5 e Au³(aq) + 3 e PbOy(s) +4 H(aq) + 2e¯ C+20 00, (aq) +14 H(aq) + 6 e 00+ 4H*(aq) + 4 MnO₂ (s) +4 H(aq) +2e¯ 10₂(aq) + 6H(aq) +5e Big(0+20 VO₂ (aq) + 2 H(aq) + e¯ NO₂ (aq) + 4H(aq) + 3 e CO₂(0)+ Ag (ag)+e Fe(aq) + e 0)+2 H(aq) +2 e Mn0, (aq) + e (8) +26 Cu (aq) + e -2 (aq) -2 H₂O(0) Which of the following is the strongest oxidizing agent? -PbSO4(s) + 2 H₂O(0) →→MnO₂ (s) + 2 H₂O(0) Mn²(aq) + 4H₂O(1) - Au(s) ·Pb²+ (aq) + 2 H₂O(0) - 2 Cr (aq) 2 Cr³+ (aq) + 7 H₂O(1) → 2 H₂O( -2(aq) -Cu(s) Mn²+ (aq) + 2 H₂O(0) ₂(aq) + 3 H₂0(0) -2 Br (aq) - Vo²+ (aq) + H₂O( -NO(g) + 2 H₂O(1) — CIO; (aq) -Ag(s) -Fe²+ (aq) -) HyOz(ag) MnO₂ (aq) 1.00 0.96 0.95 0.80 0.77 0.70 0.56 0.54 0.52 E° (V) 2.87 1.78 1.69 1.68 1.51 1.50 1.46 1.36 1.33 1.23 1.21 1.20 1.09

Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG° for the following redox reaction. Round your answer to 3 significant digits. 6CI (aq)+2Cr0 (aq)+8H,0 (1) → 3Cl, (g)+2Cr(OH), (s) + 10OH (aq)

Calculate the standard cell potential (E°cell) of a galvanic cell made of lead and cadmium. Then determine the non-standard cell potential (Ecell) if the molarity of the cadmium solution was 0.20 M, and the molarity of the lead solution was 0.10 M. The temperature of the cell is 300 K.   Cd2+(aq) + 2e–   →  Cd(s)                     E° = –0.403 V                         Pb2+(aq) + 2e–   →   Pb(s)                     E° = –0.126 V     E°cell = ______________     Ecell = ______________

The cell potential of the following cell at 25°C is 0.475 V: Zn|Zn?t (1.0 M) || H* (test solution) H2 (1.0 atm)|Pt What is the pH of the test solution? The standard reduction potential for Zn+ is –0.76 V. (Enter your answer to two significant figures.) pH =

Calculate the equilibrium constant, K, for the reaction shown at 25 °C. Fe+(aq)+B(s) + 6 H, O(1) → Fe(s) + H,BO,(s) + 3 H,O*(aq) The balanced reduction half-reactions for the equation and their respective standard reduction potential values (E") are Fe3+ (aq) + 3 e- → Fe(s) H,BO, (s) + 3 H,0*(aq) + 3 e- – B(s) + 6 H, O(1) E' = -0.04 V E' = -0.8698 V K =

Calculate Eu for the following cell: Pt(s) | H₂(g) | H *(aq) || Pb2+ (aq) | PbSO4(s) | Pb(s) given the following standard reduction potentials. 2 H(aq) + 2e →→ H₂(g) PbSO4(s) + 2e →→ Pb(s) + SO42 (aq) O a. -0.712 V O b. +0.356 V O c. -0.356 V O d. +0.712 V O e. -0.178 V E° = 0.000 V E = -0.356 V

Consider the electrolysis reactions cathode anode whose standard reduction potentials can be found in this table. Calculate the voltage needed to drive the net reaction if the current is negligible. Ecell = Cu²+ (aq, 0.177 M) +2e=Cu(s) H₂O(1) ⇒ 10₂(g, 1 bar) + 2 H+ (aq, 0.195 M) + 2 e¯ The cell has a resistance of 66.0 and a current of 7.17 mA when the cathode overpotential is 0.231 V and the anode overpotential is 0.050 V. Calculate the voltage needed to overcome these effects and drive the net reaction. Ecell = V V

Calculate the cell potential for the reaction as written at 25.00 °C, given that [Mg? +] = 0.788 M and [Fe² +] = 0.0200 M. Use the standard reduction potentials in this table. Mg(s) + Fe2+(aq) = Mg2+(aq) + Fe(s) E = V

Consider the following reduction potentials. Ag*(aq) + e → Ag(s) E° = + 0.80 V Sn2*(aq) + 2 e -→Sn(s) E° = - 0.14 V Mg2 (aq) + 2 e - → Mg(s) E° = - 2.37 V a.) Calculate the cell potential at standard conditions (E°cell) for a voltaic cell of Mg2+/Mg and Ag*/Ag, in Volts. E° cell = v b.) Calculate Ecel in V at 298 K if [Mg2*] = 0.100 M and [Ag*] = 0.100 M. %3D Ecell = V