In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of...
In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions: [Zn2+ ] = 0.021 M [H+ ]= 1.3 M partial pressure of H2 = 0.32 atm. Calculate Ecell at 298 K (enter to 3 decimal places). Zn2+ (aq) + 2e − ⟶ Zn(s) E° = − 0.76 V 2H+ (aq) + 2e − ⟶ H2(g) E° = 0.00 V We were unable...
Consider a voltaic cell al 25 degree Celsius in which the reaction is: Zn (s) + 2H+ (aq) → Zn2+ (aq) + H2(g) It is found that the voltage (think Ecell) is +0.560 V when [Zn2+1 -0.85 M and PH2 = = 0.988 atm. What is the pH in the H2-H half-cell?
a.) A voltaic cell is constructed in which the cathode is a standard hydrogen electrode and the anode is a hydrogen electrode (P(H2) = 1 atm) immersed in a solution of unknown [H+]. If the cell potential is 0.204 V, what is the pH of the unknown solution at 298 K? b.) An electrochemical cell is constructed in which a Cr3+(1.00 M)|Cr(s) half-cell is connected to an H3O+(aq)|H2(1 atm) half-cell with unknown H3O+ concentration. The measured cell voltage is 0.366...
Draw the voltaic cell that will give the most positive cell potential choosing from the following half reactions: Eo (vs. SHE) Zn2+ (aq) + 2e- Zn (s), Eo= -0.76 Al3+ (aq) + 3e- Al (s), Eo= -1.66 Cr3+ (aq) + 3e- Cr (s), Eo= -0.74 Co2+ (aq) + 2e- Co (s), Eo= -0.28
Design a voltaic cell with the following two reduction half -reactions : Design a voltaic cell with the following two reduction half-reactions Ag+(aq) + e-→ Ag(s) Zn2+(aq) + 2 e-→Zn(s) E 0.80 V Eo = 0.76 V Calculate Eocell and the equilibrium constant K for th copy of Final Exam cover sheet. e voltaic cell at 298 K. Click here for a E° cell -0.04 V and K-4.7 E cell-0.04 V and K-0.21 O Eocel =-0.04 V and K =...
A voltaic cell is set up with one beaker containing 1.0 M Zn(NO 3) 2 and a zinc electrode, and another beaker containing 1.0 M Ni(NO 3) 2 and a nickel electrode. Given the following standard reduction potentials, answer the 3 questions below: Eº Zn2+(aq) + 2e → Zn(s) -0.76 V Ni2+(aq) + 2e → Ni(s) -0.23V Part a. Write out the half-cell reaction that occurs at the anode of the voltaic cell. Part b. In which direction do...
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);...
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!
tion 3 of 40 ) A voltaic cell employs the redox reaction: Zn (s) + Pb2+ (aq) — Zn2+ (aq) + Pb () where Pb2+ (aq) + 2e Zn2+(aq) + 2e Pb(s) E = -0.1262 V Zn(s) E = -0.7618 V The cell potential (Ecell) of this voltaic cell when [Pb2+1 = 1.10 M and [Zn2+) = 0.00110 Mis:
Use the half-reactions below to produce a voltaic cell with the given standard cell potential. Standard Cell Potential Co- (aq) + e-Cot (aq) E = +1.82 V 1.53 V 2H(aq) + 2e-H2(g) E = +0.00 V Pb2+ (aq) + 2e-Pb(s) E = -0.13 V Fe (aq) + e-Fel+ (aq) E = +0.77 V Ag (aq) + e-Ag(s) E = +0.80 V Sn* (aq) + 2e Sne (aq) 20.13 V Cu- (aq) + e- Cu(aq) E = +0.15 V Zn²+ (aq)...