Calculate the value of the equilibrium constant for the reaction 2MnO4- + 3Mn2+ + 2H2O ↔ 5MnO2(s) + 4H+ using the following half-reactions: MnO4- + 4H+ + 3e- ↔ MnO2(s) + 2H2O E°=+1.692V MnO2(s) + 4H+ + 2e- ↔ Mn2+ + 2H2O E°=+1.230V

Lithium metal is a very strong reducing agent and reacts vigorously with 1 M hydrochloric acid to produce hydrogen gas. The relevant half reactions are Li+(ag) + e- = Li(s) E = -3.04 V 2H+(ag) + 2e- = H2g) E = +0.00 V - Lithium also reacts vigorously with pH-neutral iced water. Assuming a hydrogen gas pressure of 0.1 atm and a lithium ion concentration of 0.1 mol/L and assuming that the standard cell potential does not change with temperature, determine the cell potential.  - A more accurate calculation for the previous part would need to take into account the change in E° with temperature. What information would be needed to do this, and do you expect the actual cell potential would be higher or lower than that calculated? [Hint: Gibbs Free Energy change.]

The following half-reaction °E = -2,07 V. AlF63-(aq) + 3e- à Al(s) + 6F-(aq) calculate the equilibrium constant at 25°C for this reaction: Al3+(aq) + 6F-(aq) ↔ AlF63-(aq)                    K = ?

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Given the following electrochemical cell, calculate the potential for the cell in which the concentration of Ag+ is 0.0765 M, the pH of the H* cell is 2.700, and the pressure for H₂ is held constant at 1 atm. The temperature is held constant at 55°C. Ag Ag+ H+ Standard reduction potentials: 2H+ +2e¯ → H₂ Eº = 0 V Ag+ le - H₂(g) → Ag Eº=0.800 V

What is the pH in the cathode for the following galvanic cell made from the following two half reactions? The initial concentration of gold ion is 1.1x10-3 M, the initial concentration of permanganate is 6.42 M, and the Ecell is 0.15 V. Assume the temperature is 298K. Au3+(aq) + 3e- →   Au(s)                                     E° = 1.52 VMnO4-(aq) + 4H+(aq) 3e- → MnO2(s) + 2H2O(l)      E° = 1.70 V

Examine the following half-reactions and select the strongest reducing agent among the species listed.      AuBr4(aq) + 3e– → Au(s) + 4Br–(aq)                           E° = 0.854 V      Mn2+(aq) + 2e– → Mn(s)                                             E° = –1.185 V      K+(aq) + e– → K(s)                                                       E° = –2.931 V      F2O(aq) + 2H+(aq) + 4e– → 2F–(aq) + H2O(l)           E° = 2.153 V Group of answer choices K K+ F2O F–

Determine E°, AG°, and K for the overall reaction from the balanced half-reactions and their standard reduction potentials. 4 Fe3+ + 2 H,0 4 Fe²+ + O, + 4H+ Fe3+ + e-= Fe²+ E° = 0.771 V 0, + 2 H* + 2 e¯2 H,0 E° = 1.229 V E° = V K = Is the overall reaction spontaneous as written? AG° = spontaneous not spontaneous

1) Calculate the following for the reaction at 298.0 K . The reaction is 2 SmCl3(aq) + 6 LiCl(aq) + 6 e- → 2 Sm(s) + 6 Li+(aq) + 6 Cl-(aq) A) The standard Gibbs free energy change of the reaction @ 298.0 K B) The equilibrium constant for the reaction C) The standard cell potential 2) Calculate the free energy change and the cell potential for the reaction at 298.0 K if all reactants are at 2.00 M and/or 2.00 atm and all products are at 0.0100 M and/or 0.0100 atm

2+ 4+ a. Will Fe+ oxidize Sn to Sn*t ? Assume that the reaction occurs at standard electrochemical conditions. Half-Reaction e° (V) Fes+ +e → Fe2+ +0.77 Sn n4+ + 2e → Sn²+ +0.15 O yes O no b. Will dichromate ions oxidize chloride ions to free chlorine in acidic solution? Assume that the reaction occurs at standard electrochemical conditions. Half-Reaction e° (V) Cr2 0, + 14H+ + 6e¯ → 2Cr³+ + 7H2O +1.33 Cl2 + 2e → 2C1 +1.36 O yes O no