3. Show your calculation for the potential (E) of the voltaic cell if [Mn) 0.11M instead...
please help and answer all questions 4. What is the cell potential for Mn Mn2||Co2|Co, based on the standard potentials? (2 pts) Mn(s) +Co+2 (aq) → Co(s) + Mn+2(aq) 5. A Student constructed a Mg Mg2||Ag|Ag cell and measured a cell potential of 3.17 V. Calculate the cell potential of the Mg Mg2 half reaction. (2 pts) Mg(s) + 2Ag (aq) - 2Ag(s) + Mg(aq) 6. Will silver metal react spontaneously with HCl(aq) to produce H:()? Explain. (2 pts) 2Ag(8)...
2 A voltaic cell is set up with one beaker containing 1.0 M Cu(NO 3) 2 and a copper electrode, and another beaker containing 1.0 M Mn(NO 3) and a manganese electrode. Given the following standard reduction potentials, answer the 3 questions below: E Cu2+(aq) + 2e Cu(s) 0.34 V Mn2+(aq) + 2e + Mn(s) 1.18V Part a. Write out the half-cell reaction that occurs at the anode of the voltaic cell. Part b. In which direction do electrons flow?...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 oC. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M, respectively. What is the initial cell potential? Hint: the standard potential of Pb2+ + 2e- → Pb(s) is -0.130 and the standard potential of Cu2+ + 2e- → Cu(s) is +0.340. Hint #2: Use [Cu2+] as the product and [Pb2+] as the reactant. The equation we are supposed to use is E= Eo-...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 oC. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M, respectively. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V? Cu2+(ag) + Pb(s) → Cu(s) + Pb2+(ag) Hint: [Pb2+] + [Cu2+] = (1.5M + 0.05M) = 1.55 M total Hint: the standard potential of Pb2+ + 2e- → Pb(s) is -0.130 and the standard potential...
Enter electrons as e A voltaic cell is constructed in which the anode is a Mn Mn+ half cell and the cathode is a Ag Agt half cell. The half-cell compartments are connected by a salt bridge. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) The anode reaction is: The cathode reaction is: The net cell reaction is: In the external circuit,...
26: Voltaic Cell Diagram, Cell Potential, free-energy, K Time Left:0:14:34 Rashad Reed: Attempt 1 The following reaction is occurring in an voltaic cell at 25 °C Mn (s) + Br2 (1) ► Mn2+ (aq) + 2 Br(aq) The following are the standard reduction potentials: Mn2+ (aq) + 2e + Mn (5) E' = -1.18 V Br2 (1) + 2 e + 2 Br- (aq) E = 1.07 V On the paper you will submit, sketch the voltaic cell making sure...
Calculate the theoretical cell potential (E°) of a galvanic cell under standard conditions made up of copper and magnesium (see Part II and Table 1 for more information). PARTIL Creating and Testing Voltaic Cells Introduction and Background for the Voltaic Cells A galvanic cell (sometimes more appropriately called a voltaic cell) consists of two half-cells joined by a salt bridge that allow ions to pass between the two sides in order to maintain electroneutrality. Each half-cell contains the Components of...
A voltaic cell employs the following redox reaction: Sn2+(aq)+Mn(s)---- Sn(s)+Mn2+(aq) Calculate the cell potential of 25 degrees Celsius under each of the following conditions. Part A: Sn2+= 1.15*10^-2 M; and Mn2+= 2.37 M Part B: Sn2+= 2.37 M; and Mn2+= 1.15*10^-2
A voltaic cell employs the following redox reaction: Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq) Calculate the cell potential at 25 ∘C∘C under each of the following conditions. Part A [Sn2+]= 1.34×10−2 MM ; [Mn2+]= 2.51 MM . Express your answer using two significant figures. Part B [Sn2+]=2.51 MM ; [Mn2+]=1.34×10−2 MM .
Calculate the cell potential, the equilibrium constant, and the free-energy change for: Ca(s)+Mn2+(aq)(1M)⇌Ca2+(aq)(1M)+Mn(s) given the following Eo values: Ca2+(aq)+2e−→Ca(s) Eo = -1.59 V Mn2+(aq)+2e−→Mn(s) Eo = -0.54 V 1.) Calculate the equilibrium constant. 2.) Free-energy change?