A galvanic cell is based on the following half-reactions: Fe2+ + 2e-→ Fe(s) Eº = -0.440 V 2H+ + 2e-→ H2(g) Eº = 0.000 V where the iron compartment contains an iron electrode and [Fe2+] = 1.00 × 10-3 M and the hydrogen compartment contains a platinum electrode, PH2 = 1.00 atm , and a weak acid, HA, at an initial concentration of 1.00 M. If the observed cell potential is 0.320 V at 25ºC, calculate the Ka value for the weak acid HA.
Ka = ?
A galvanic cell is based on the following half-reactions: Fe2+ + 2e-→ Fe(s) Eº = -0.440...
Need help with pS Calculate Eº for the half-reaction, FeS(s) +2e=Fe(s) + S2-(aq) given that the solubility product for FeS at 25°C is 8.0 x 10-19 and the standard reduction potential of the half-reaction Fe? + (aq) +2e = Fe(s) is -0.440 V. -0.974 Fes- Construct the line notation of a cell that consists of a saturated calomel reference electrode, the FeS half-reaction, and an iron indicator electrode, which can be used to measure the s2-concentration of the solution. HGO)...
HIU Blud cell potential of a galvanic cell based on the following reduction half- reactions at 25 °C Cd + 2e → Cd Eº - -0.403 V Pb2+2e → Pb Eº - -0.126 V where (Cd-) - 0.040 M and (Pb) - 0.400M (13.) A zinc electrode is submerged in an acidic 0.40 M Zn?' solution which is connected by a salt bridge to a 1.50 M Ag' solution containing a silver electrode. Determine the initial voltage of the cell...
Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe (s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn** (aq) +2Cr(s) → 2Cr** (aq) +3Sn²+ (aq) 3. What is the cell potential, Ecell, for the reaction above if [Sn] = 1.00 M, [Cr3+1 = 0.0200 M and [Sn2+] = 0.0100 M?
Two half-reactions are shown with their standard cell potentials. If a galvanic cell is constructed using them, which electrode would be the anode, and what would the cell potential be? Fe2+ (aq) + 2e → Fe(s); E=-0.44 V Ag+(aq) + 1e → Ag(s); E=0.80 V O a. Iron, 1.24 V O b. Silver, 0.36 V O c. Iron, 0.36 V O d. Iron, -0.36 V O e. Silver, -0.36 V
Consider a galvanic cell based on the following two half reactions under standard conditions at 298 K: Fe2+ + 2 e → Fe -0.440 V Cd2+ + 2 e → Cd -0.403 V What will the potential of the cell be when the cathode solution concentration changes by 0.657 M? Cell potential =
Table 20.2 Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr(s) -0.74 Fe2+ (aq) + 2e- Fe (5) -0.440 Fe3+ (aq) + e - Fe2+ (s) +0.771 Sn4+ (aq) + 2e- Sn2+ (aq) +0.154 The standard cell potential (Eºcell) for the voltaic cell based on the reaction below is V. 35n4+ (aq) + 2Cr (s) → 2Cr3+ (aq) + 3Sn2+ (aq) A) +1.94 B) +0.89 C) +2.53 D) -0.59 E) -1.02
2. A standard state galvanic cell is constructed using the following half-reactions. Fe3+ (aq) + Fe2(aq) Fe2(aq) + 2e → Fe(s) a. Fill in the cell diagram. Label the anode, cathode, the reactants and products in each solu direction of electron flow through the wire, and ion flow through the salt bridge. (5 pts) KNO, Salt Bridge (+) electrode (-) electrode b. Calculate the concentration of each ion in the cell at equilibrium. (10 pts)
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...
Half-reaction Cr3+ (aq) + 3e--Cr(s) Fe2+ (aq) + 2e - Fe(s) Fe3+ (aq) - Fe2+ (5) Sn+ (aq) + 2e - Sn2(aq) E (V) -0.74 -0.440 +0.771 +0.154 1. Calculate the standard cell potential for the voltaic cell based on the reaction below, given the table above: 35nt(s) + 2Cr (s) - 2 C (s) + 3 Sn () ANSWER: 2. Calculate the standard cell potential for the voltaic cell based on the reaction below, Riven the table above 3Feb...
Answer all question (a-d) thank you Consider a galvanic cell which has an Iron electrode submerged in a Fe(NO3)2 soln. and a Gold electrode submerged in a Au(NO3)3 soln. a) Find the Ecell of this cell under standard conditions: (2pts) Ans: Au3+ (aq) + 3e = Aus) ER 1.50 V + 2e = Fe(s) Er = -0.44 V Fe2+ (aq) b) Write out the balanced net reaction of this cell? (3pts) c) What is the cell potential if the concentration...