2Fe3+(aq)+H2(g)→2Fe2+(aq)+2H+(aq).
What is the emf for this cell when [Fe3+]= 3.90 M , PH2= 0.94 atm , [Fe2+]= 1.5×10−3 M , and the pH in both compartments is 3.80?
2Fe3+(aq)+H2(g)→2Fe2+(aq)+2H+(aq). What is the emf for this cell when [Fe3+]= 3.90 M , PH2= 0.94 atm...
A voltaic cell utilizes the following reaction: 2Fe3+(aq)+H2(g)→2Fe2+(aq)+2H+(aq). emf of this cell under standard conditions E∘ = 0.771 V What is the emf for this cell when [Fe3+]= 3.70 M , PH2= 0.95 atm , [Fe2+]= 1.0×10−3 M , and the pH in both compartments is 3.95? Express your answer using two significant figures.
1. A voltaic cell is constructed that is based on the following reaction: Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq). A. If the concentration of Sn2+ in the cathode compartment is 1.30 M and the cell generates an emf of 0.25 V , what is the concentration of Pb2+ in the anode compartment? B. If the anode compartment contains [SO2−4]= 1.30 M in equilibrium with PbSO4(s), what is the Ksp of PbSO4? 2. A voltaic cell utilizes the following reaction: 2Fe3+(aq)+H2(g)→2Fe2+(aq)+2H+(aq). A. What is the emf...
Question 1 1 pts For the cell: Zn(s) + 2H+ (aq) + Zn2+(aq) + H2 (g) If [Zn2+) = 1 M, PH2 = 1 atm, and E = 0.549 V what is pH? Question 2 1 pts Calculate the voltage for the following cell at 25° ZnZn+2 ( 2.832 M) || Cd2+ (0.027 M)| Cd
1)A voltaic cell operates at 298 K according to the following reaction: 4 Fe2+(aq) + O2 (g) + 4 H+(aq) → 4 Fe3+ (aq) + 2 H2O (l) What is the emf of this cell when [Fe2+] = 6.5908E-4 M, [Fe3+] = 0.699 M, pressure O2 = 0.540 atm and pH = 3.10? 2)A voltaic cell operates at 298 K according to the following reaction: 3 Fe2+(aq) → Fe (s) + 2 Fe3+ (aq) What is the emf of this...
An electrochemical cell consists of a Pt|H+(aq,1.00 M)|H2(g) cathode connected to a Pt|H+(aq)|H2(g) anode in which the H+ concentration is that of a buffer consisting of a weak acid, HA(0.136 M), mixed with its conjugate base, A-(0.142 M). The measured cell voltage is E°cell = 0.196 V at 25 °C, with PH2 = 1.00 atm at both electrodes. Calculate the pH in the buffer solution and the Ka of the weak acid. pH = _______ Ka = _______
An electrochemical cell consists of a Pt|H+(aq,1.00 M)|H2(g) cathode connected to a Pt|H+(aq)|H2(g) anode in which the H+ concentration is that of a buffer consisting of a weak acid, HA(0.160 M), mixed with its conjugate base, A-(0.120 M). The measured cell voltage is E°cell = 0.228 V at 25 °C, with PH2 = 1.00 atm at both electrodes. Calculate the pH in the buffer solution and the Ka of the weak acid. pH = Ka =
1.) Given the following notation for an electrochemical cell Pt(s) | H2(g) | H+(aq) || Ag+(aq) | Ag(s), what is the balanced overall (net) cell reaction? A. H2(g) + 2Ag(s) ® H+(aq) + 2Ag+(aq B. H2(g) + 2Ag+(aq) ® 2H+(aq) + 2Ag(s) C. H2(g) + Ag+(aq) ® H+(aq) + Ag(s D. 2H+(aq) + 2Ag(s) ® H2(g) + 2Ag+(aq) E. 2H+(aq) + 2Ag+(aq) ® H2(g) + 2Ag(s) 2.) Calculate E°cell for the following (nonspontaneous) reaction: Cd(s) + 2Fe3+(aq) ® 2Fe2+(aq) + Cd2+(aq) → A. -0.37...
Consider the cell Pt(s)|H2(g,1atm)|H+(aq,a=1)|Fe3+(aq),Fe2+(aq)|Pt(s) given that Fe3++e−⇌Fe2+ and E∘=0.771V at 298.15 K. If the cell potential is 0.683 V, what is the ratio of Fe2+(aq) to Fe3+(aq)? What is the ratio of these concentrations if the cell potential is 0.807 V?
8) An electrochemical cell consists of a Pt|H+(aq,1.00 M)|H2(g) cathode connected to a Pt|H+(aq)|H2(g) anode in which the H+concentration is that of a buffer consisting of a weak acid, HA(0.115 M), mixed with its conjugate base, A-(0.192 M). The measured cell voltage is E°cell = 0.168 V at 25 °C, with PH2 = 1.00 atm at both electrodes. Calculate the pH in the buffer solution and the Ka of the weak acid. pH = Ka =
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);...