3. (5 pts each) Given the following reactions, determine the following Ni2+ + 2e - →...
Given the following information: Ni2+ + 2e- Ni(s) Eº = -0.25 V Ag* + e. Ag(s) E° = 0.80 V 1) Determine the cell potential of the spontaneous redox reaction. 2) Give the net ionic equation for the overall balanced redox reaction. 3) Label the Galvanic Cell below to model the reaction above. Please include: • substances in each compartment • the salt bridge • the anode • the cathode 4) Give the line notation representation of this galvanic cell....
Selective Reduction The standard reduction potential for the half-reaction: Sn4+ + 2e - Sn2+ is +0.15 V. Consider data from the table of standard reduction potentials for common half-reactions, in your text. For a galvanic cell under standard conditions, which of the following anodic half reactions would produce, at the cathode a spontaneous reduction of Sn4+ to Sn2+ but not Sn2+ to Sn. no yes yes yes yes yes Fe — Fe2+ + 2e- Sn2+ Sn4+ + 2e- Sn Sn2+...
Consider the following electrochemical cell at room temperature. 1.05 V N м Salt Bridge Ni2+an (110 The half cell and the reuction potential for the anode is: Half-Reaction E (V) Ni2+ (aq) + 2e ® Ni(s) -0.25 Using this information from the diagram, determine the standard reduction potential of metal M. +0.25 V O-1.40V +0.80 v -0.80 V +1.40 V
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.26 M )+2e- Red: MnO-4(aq, 1.50 M )+4H+(aq, 2.7 M )+3e-→MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C.
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...
19 20 Question 16 Half-cell Potentials: Half Reaction: E' value + 0.80 V +0.77 V Agte → AS Fe3+ + + Fe2+ Cu2+ 2e → Cu Pb2+ + 2e → Pb +0.34 V -0.13 V Ni2+ + 2e → NI -0.25 V - 0.40 V Cd2+ +2e → ca Fe2+ + 2e → Fe Zn2+ + 2e → Zn -0.44 V - 0.76 V A13+ +3 → AI - 1.66 V Consider an electrochemical cell constructed from the following half...
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 Part A Part complete An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.017M)+2H+(aq,1.3M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l) Calculate the cell potential under these nonstandard concentrations.
Selective Oxidation The standard reduction potential for the half-reaction Sn4+ + 2e - Sn2+ is +0.15 V. Consider data from the table of standard reduction potentials for common half-reactions, in your text. For a galvanic cell under standard conditions, which of the following cathodic half reactions would produce, at the anode, a spontaneous oxidation of Sn to Sn2+ but not Sn2+ to Sn4+. 2H+ + 2e - H2 Fe3+ + 3e + Fe Sn2+ + 2e Fe2+ + 2e →...
EXPERIMENT 1: List the measured potential for Cell 3: Zn|Zn(NO3)2 || Pb(NO3)2|Pb. Based on your observations, do you expect given electrochemical cell to be spontaneous or nonspontaneous? Explain your answer. Pb|Pb(NO3)2||Zn|Zn(NO3)2 EXPERIMENT 1: What would happen to the measured cell potentials if 30 mL solution was used in each half-cell instead of 25 mL? EXPERIMENT 1: Calculate the theoretical standard cell potential for the electrochemical cell that includes the reaction. Mn+Pb2+⟶Mn2++Pb The standard reduction potentials for each half reaction are...
calculate ecell for the following electrochemical cell at 25 C. Pt(s)| H2 (g,1.00 atm) | H+ (aq, 1.00 M) || (Pb2+ (aq, 0.150 M) | Pb (s) Pb^2+(aq) + 2e- ---> Pb (s) Eo=-0.126 V 2H^+ (aq) + 2e- ---> H2 (g) Eo=0.00 V Thank you!