For the given cell reaction
Cathode reaction: Zn2+ (aq)
+ 2e Zn (s) ;
EoZn2+/Zn = - 0.76 V.
Anode reaction: Pb (s) - 2e Pb2+
(aq) ;
EoPb2+/Pb = - 0.13
V
Eocell = Eocathode - Eoanode = - 0.76 - (-0.13) = - 0.63 V.
Question 7 What is the value (in V) of Eºcell for the following reaction? Zn2+ (aq)...
- Zn2+ Choose... Choose... (1pts) Identify the complete redox reaction for a ZnZn2+1|Cu2+1Cu cell. A. Zn(s) + Cu?+ (aq) (aq) + Cu(s) B. Zn(s) + Cu(s) → Zn2+ (aq) + Cu2+ (aq) C. Zn2+ (aq) + Cu(s) Zn(s) + Cu2+ (aq) D. Zn (s) + 2 Cu(s) — Zn2+ (aq) + 2 Cu2+ (aq) (1pts) Identify the complete redox reaction for a Zn/Zn2+||Pb2+1Pb cell. A. Zn (s) + Pb(s) Zn2+ (aq) + Pb2+ (aq) B. Zn2+ (aq) + Pb(s) Zn(s)...
For the reaction, Zn(s) + Pb2+(aq) <--> Zn2+(aq) + Pb(s) You would attach the black probe to...
What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Pb2+ concentration is 1.38 M and the Zn2+ concentration is 9.41*10-4 M? Pb2+(aq) + Zn(s) —— Pb(s) + → Pb(s) + Zn2+(aq) Answer: V The cell reaction as written above is spontaneous for the concentrations given:
For the reaction, Pb(s) + 2HCIO(aq) + 2H+(aq) → Pb2+(aq) + Cl2(g) + 2H2O(1) the value of Eºcell is 1.74 V. What is the value of Eºcell for the related reaction given below? YPb2+(aq) + 42Cl2(g) + H2O(1) → YPb(s) + HCIO(aq) + H+(aq) O 1.74 V -0.87 V -1.74 V 0.87 V 0.0 V
Consider the cell represented by the following line notation: Zn(s) | Zn2+ (aq) || ClO2 (aq)| ClO2(g) | Pt(s) Assuming standard conditions, calculate Eºcell: Zn2+ (aq) + 2e - Zn(s) E' = -0.76 V ClO2(g) + e - ClO2 (aq) F° = 0.95 V Express your answer to two decimal places. Eºcell
Question 7 (1 point) Standard reduction potentials for the Zn2+/Zn and Pb2+/Pb couples are -0.76 and - 0.13 V, respectively. The galvanic cell below uses the half-cells Pb2+[Pb and Zn2+|Zn, and a salt bridge containing KCl(aq). The voltmeter gives a positive voltage reading. voltmeter salt bridge CD-> The identities of B and D, respectively, are O Pb and Cl- Ozn and Cl- O Pb and K+ O Zn and K+
1. What is the value (in V) of Eocell for the following reaction? Co2+ (aq) + Be (s) → Co (s) + Be2+ (aq) 2.What is Eocell (in V) for a redox reaction where one electron is transferred with an equilibrium constant (K) of 1.44 x 10-14? 3.Consider the following reaction: Cu2+ (aq) + Pb (s) → Cu (s) + Pb2+ (aq) What will be Ecell for this reaction (in V) when [Cu2+] = 0.500 M and [Pb2+] = 0.0350...
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:
Using the standard reduction potentials given below, choose the reaction than can only be achieved through electrolysis. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V Pb2+(aq) + 2e + Pb(s) E° = -0.13 V Fe2+(aq) + 2e Fe(s) E° = -0.44 V Zn2+(aq) + 2e + Zn(s) E° = -0.77 V Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq) o Fe2+(aq) + Zn(s) → Fe(s) + Zn2+(aq) Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) Cu2+(aq) + Fe(s) → Cu(s) +...
Use the References to access important values if needed for this question. What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Pb2+ concentration is 1.31 M and the Zn2+ concentration is 3.17*10* M? Pb2+ (aq) + Zn(s) Pb(s) + Zn2+ (aq) Answer: V The cell reaction as written above is spontaneous for the concentrations given: Submit Answer Retry Entire Group 4 more group attempts remaining