3) Calculate the standard cell potential E ^o and explaination for 4 please :) o Problem...
2. Calculate the standard cell potential for each of the following redox reactions, and then predict whether each will occur spontaneously as written. a. Sr(s) + Fe2+(aq) → Sr2+ (aq) + Fe(s) b. 2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd(s) 3. Calculate the standard cell potential, Eºcell, for each of the voltaic cells in Part II of the experiment. a. Zn(s) | Zn2+ (aq, 1.0 M) || Cu2+ (aq, 1.0 M) Cu(s) b. Zn(s) | Zn2+(aq, 1.0 M)...
For each set of materials below:
i.Write the cell notation (line notation)for a galvanic cell
and
ii.Calculate the standard cell potential
a. Pb(s), Zn(s), Pb2+(aq),Zn2+(aq)
b. Ag(s), Fe(s), Ag+(aq),Fe2+(aq)
c. Mg(s), Zn(s), Mg2+(aq),Zn2+(aq)
d. Fe(s), Mg(s), Fe2+(aq), Mg2+(aq
e. Ag(s), Pb(s), Pb2+(aq), Ag+(aq)
2. For each set of materials below: Write the cell notation (line notation) for a galvanic cell and Calculate the standard cell potential ii. a. Pb (). Zn (3), Pb2+(aq), Zn2+ (aq) 6. Ag (3), Fe (3),...
Calculate the cell potential for the reaction as written at 25.00°C, given that [Zn2+] = 0.758 M and [Fe2+] = 0.0140 M. Use the standard reduction potentials in this table. Zn(s) + Fe2+ (aq) = Zn²+ (aq) + Fe(s) Zn2+(aq) + 2e- → Zn(s) -0.76 Fe2+(aq) + 2e- > Fe(s) -0.44
8. The standard cell potential (E°cell) for the reaction below is +1.10 V. Calculate the cell potential for this reaction when the concentration of [Cu2+] = 1.0 × 10-5 M and [Zn2+] = 3.5 M. Zn (s) + Cu2+ (aq) → Cu (s) + Zn2+ (aq)
Calculate the standard cell potential and determine if the reaction is spontaneous in the forward direction(as written). Identify each oxidizing agent and reducing agent. Ni (s) + Zn2+(aq) ----> Ni2+(aq) + Zn (s) Ni (s) + Pb2+(aq) -------> Ni2+(aq) + Pb (s) Al (s) + 3 Ag+(aq) ---------> Al3+(aq) + Ag (s) Pb (s) + Mn2+(aq) ----------> Pb2+(aq) + Mn (s)
Cell Potential and Equilibrium Standard reduction potentials The equilibrium constant, K, for a redox reaction is related to the standard cell potential, Ecel, by the equation Reduction half-reaction (V) Ag+ (aq) + e-→Ag(s) Cu2+ (aq) + 2e-→Cu(s) 0.34 Sn (a) 4e-Sn(s 0.15 2H' (aq) + 2e-→H2 (g) Ni2+ (aq) + 2e-→Ni(s)-0.26 Fe2+ (aq) + 2e-→Fe(s)-0.45 Zn2+ (aq) + 2e-→Zn(s)-0.76 Al3+ (aq) +3e-→Al(s) -1.66 Mg2+ (aq) + 2e-→Mg(s) -2.37 0.80 n FEcell where n is the number of moles of electrons...
What is the standard cell potential of a cell created using a cadmium electrode in 0.45 M Cd(NO3)2(aq) and a graphite electrode in an aqueous solution that is 0.45 M in both Fe2+ and Fe3+? 0.38 V 1.95 V 1.94 V 1.17 V 1.18 V What is the standard cell potential of a cell which utilizes the following reaction: Zn (s) + Ni2+ (aq) → Ni (s) + Zn2+ (aq) All solutions are 1.0 M and the reaction occurs at...
I need help with questione 1-12 and discussion question 1 and
2. The previous pictures help determine the chart. Please Show Work
thank you so much
An oxidation half-reaction is characterized by electrons appearing on the product side. The oxidation of aluminum for instance would be represented thusly: Al(s) → Al3+ + 3e- (1) An reduction half-reaction is characterized by electrons appearing on the reactant side. The reduction of ferrous iron for instance would be represented thusly: Fe2+ + 2e...
Calculate the cell potential, E, for the given reactions at 25.00 °C using the ion concentrations provided. Then, determine if the cells are spontaneous or nonspontaneous as written. Refer to the table of standard reduction potentials (E Pt(s)Fe2+(aq) Pt2+(aq) + Fe(s) [Fe2+ [Pt2+] = 0.0023 M = 0.011 M The cell is V E = O not spontaneous. O spontaneous Cu(s)2 Ag (aq) Cu2+(aq) + 2 Ag(s) [Cu2+0.031 M [Ag*] = 0.031 M The cell is V E = O...
Consider the following redox reactions. For each reaction, calculate the standard cell potential. See Table 12-2 from your book for a list of standard reduction potentials (or lecture notes CH12). A. Ag+ + Fe2+à Fe3+ + Ag(s) B. Zn2+ + Ni(s) à Ni2+ + Zn(s) C. 2Al3+ + 3Cu(s) à 2Al(s) + 3Cu2+