True or false ? 1. An exothermic reaction that results in a decrease of entropy for...
An electrochemical cell based on the following reaction has astandard cell voltage (Eocell) of 0.48 V Sn (s) + Cu2+ (aq) ---> Sn2+ (aq) + Cu (s) What is the standard reduction potential of tin? Sn2+ (aq) +2e- ---> Sn(s) a. -0.14 V b. 0.14 V c. -0.82 V d. 0.82 V e. none of the above
Given the following standard reduction potentials choose the cell which will work as a voltaic cell. All cells below are written according to the usual cell diagram convention. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V 2H+(aq) + 2e → H2(g) E° = 0.00 V Sn2+ (aq) + 2e → Sn(s) E° = -0.14 V Ni2+(aq) + 2e → Ni(s) E° = -0.26 V Cd2+(aq) + 2e → → Cd(s) E° = -0.40 V Sn(s) | Sn2+(aq) || Ni2+(aq)...
Find the best combination of half-cell pair from the following list, which will give the highest voltage. What is the voltage for that Galvanic cell? Given that Reduction Half-reaction Standard Potential (Eredo) Zn2+(aq) + 2e– → Zn(s) -0.763 (V) Fe2+(aq) + 2e– → Fe(s) -0.44 (V) Cu2+(aq) + 2e– → Cu(s) +0.34 (V) Sn2+(aq) + 2e– → Sn(s) -0.14 (V) Cu2+(aq) + e– → Cu+(aq) + 0.153 (V) Ag+(aq) + e– → Ag(s) + 0.80 (V) Cu+(aq) + e– →...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery that... Reduction Half-Reaction F2(g) + 2e →2F(aq) S2082 (aq) + 2e- → 25042 (aq) O2(g) + 4H+ (aq) + 4e → 2H2O(1) Br2(1) + 2e + 2Br (aq) Agt(aq) + e → Ag(s) Fe3+ (aq) + e- → Fe2+ (aq) 126) + 2e → 21 (aq) Cu2+ (aq) + 2e → Cu(s) Sn4+ (aq) + 2e → Sn2+ (aq) S(s) + 2H+ (aq) +...
Question 1) What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Pb(s) and Sn2+(aq) to give Sn(s) and Pb2+(aq) Use the reduction potential values for Sn2+(aq) of -0.14 V and for Pb2+(aq) of -0.13 V Question 2) What is the value of ΔGo in kJ at 25 oC for the reaction between the pair: Cr(s) and Cu2+(aq) to give Cu(s) and Cr3+(aq) Use the reduction potentials for Cr3+(aq) is -0.74 V and...
It is expected that a chemical reaction will occur when copper metal is combined with aqueous zinc sulfate. Explain why there will be no reaction when zinc metal and aqueous copper sulfate solution are combined. Identify the anode and the cathode, assuming a voltaic cell is constructed. Note: Be careful in the calculation of the standard cell potential ( Eo cathode - Eo anode). Do not change the sign of the given reduction potential. The sign is already taken care...
+ + + + + + + + + Metal Oxidation Reaction Lithium Li(s) - Li+ (aq) Potassium K(s) K+ (aq) Barium Ba(s) Ba2+ (aq) Calcium Ca(s) Ca2+ (aq) Sodium Na(s) Na+ (aq) Magnesium Mg(s) Mg2+ (aq) Aluminum Al(s) A13+(aq) Manganese Mn(s) Mn2+ (aq) Zn(s) Zn2+ (aq) Chromium Cr(s) Cr3+ (aq) Iron Fe(s) Fe2+ (aq) Cobalt Co(s) Co2+ (aq) Nickel Ni(s) Ni2+ (aq) Tin Sn(s) Sn2+ (aq) Lead Pb(s) Hydrogen H2(8) Copper Cu(s) Cu2+ (aq) Silver Ag(s) Ag+ (aq) Mercury...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
C-1. DETERMINE THE E° FOR A VOLTAIC CELL For each cell write the anode half reaction and the cathode half reaction. From the measured voltages, calculate the half-cell potentials for the lead and tin half-cells and the equilibrium constants for these two reactions. In these calculations, use E° = 0.34 V for the Cu2+/Cu couple. Cell trial 1 trial 2 trial 3 Cu-Pb 0.505 V 0.506 V 0.508 V Cu-Sn 0.688 V 0.688 V 0.691 V Pb-Sn 0.190 V 0.190...
2. Indicate True or False for the following scenarios and explain if the reaction occurs using “higher than" and "lower than" using the sample table 3. True/False Scenario Justification Silver metal will dissolve in nitric acid, liberating H2 gas. Chromium metal can dissolve in dilute HCI. Oxygen in moist air can oxidize Fe2+ to Fe3+ Tin metal will reduce Co2+, but not Sn4+ 178 Page (intel V IA "ost Lab Questions: To receive full credit, you must SHOW ALL YOUR...