According to Nernst equation
where n = number of electrons, F is Faraday constant and E is the voltage
The unit of Faraday constant is C/mol (coulomb/mol) and E is V (volt)
PbO2 + 4 H+ (aq) + 2e- Pb2+(aq) + 2 H2O eq. 1.1
Pb2+ (aq) + 2 e- Pb(S) eq 2.2
Taking sum of the equations 1.1 and 1.2 we get
PbO2 + 4 H+ (aq) + Pb2+ (aq) + 2 e- + 2e- Pb2+(aq) + Pb(S) + 2 H2O eq 1.3
The reduction reaction is equation 1.4 as follows
PbO2 + 4 H+ (aq) + 4e- Pb(S) + 2 H2O eq 1.4
The E0red for the reduction is = = +0.66 V
Cl2 (g) + 2e- 2 Cl- (aq) eq 1.5 E0red = +1.36 V
The equation for the oxidation is as follows (eq 1.6)
2 Cl- (aq) Cl2(g) + 2e- eq 1.6 E0ox = -1.36 V
The number of electrons for the reduction must be same for the oxidation. Thus,
4 Cl- (aq) 2 Cl2(g) + 4e- eq 1.7 E0ox = -1.36 V
Taking sum of the equations 1.4 and 1.7;
PbO2 + 4 H+ (aq) + 4e- Pb(S) + 2 H2O eq 1.4 E0red = +0.66 V
4 Cl- (aq) 2 Cl2(g) + 4e- eq 1.7 E0ox = -1.36 V
we get equation 1.8 (answer to part B);
PbO2 + 4 H+ + 4 Cl- (aq) Pb(S) + 2 H2O + 2 Cl2(g)
Thus E0cell = = +0.70 V (answer to part C)
Scores A Review | Constants | Periodic Table e Text Standard Reduction Half-Cell Potentials at 25°C...
Standard reduction half-cell potentials at 25°C E (V) E (V) 1.50 -0.45 0.80 -0.50 0.77 -0.73 0.52 -0.76 0.34 -1.18 Half-reaction Aut (aq) + 3e +Au(s) Ag+ (aq) + +Ag(s) Fe3+ (aq) +34 Fo+ (aq) Cut(aq) + Cu(s) Cu²+ (aq) + 2e +Cu(s) 2H+ (aq) - 2e +H2 (6) Fe3+ (aq) + 3e Fe(s) Pb2+ (aq) + 2e →Pb(s) Sn-(aq) + 2e +Sn(s) Ni2+ (aq) + 2e →Ni(s) Co2(aq) +2e + Co(s) ca? (aq) + 2e +Cd(s) 0.00 Half-reaction Fe(aq)...
Calculate the equilibrium constant for each of the reactions at 25 ∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Pb2+(aq)+2e− →Pb(s) -0.13 Zn2+(aq)+2e− →Zn(s) -0.76 Br2(l)+2e− →2Br−(aq) 1.09 Cl2(g)+2e− →2Cl−(aq) 1.36 MnO2(s)+4H+(aq)+2e− →Mn2+(aq)+2H2O(l) 1.21 Pb2+(aq)+2e− →Pb(s) -0.13 Br2(l)+2Cl−(aq)→2Br−(aq)+Cl2(g) Express your answer using two significant figures.
Exercise 19.65 Use tabulated half-cell potentials to calculate AG for each of the following reactions at 25 C. Part A Pb2+ (aq) +Mg(e)-Pb(a) +Mg (aq) VAZD AGn Submit Request Answer Part B Br ()+2Cl (aq)-+2Br (aq) + Cl( Express your answer using two significant figures AG Submit Request Answer Part C MnO2(a)+ 4H (aq) + Cu()Mn (ag) +2H,O( Express your answer using two significant figures
MI Review | Constants | Periodic Table Calculate the standard cell potential for each of the following electrochemical cells. Standard Electrode Potentials at 25 °C Part A Cd2+ (aq) + Mg(s)+Cd(s) + Mg2+ (aq) Express your answer using two decimal places and include the appropriate units. -2.37 Reduction Half-Reaction Cd2+ (aq) +2 e Mg2+(aq) +2 € 2 H+ (aq) +2 € Fe2+ (aq) +20 Cu2+ (aq) +2 € NO3- (aq) + 4 H+ (aq) + 3 e E°(V) + Cd(s)...
tandard reduction half-cell potentials at 25 ∘ C Half-reaction E ∘ ( V ) Half-reaction E ∘ ( V ) A u 3+ (aq)+3 e − →Au(s) 1.50 F e 2+ (aq)+2 e − →Fe(s) − 0.45 A g + (aq)+ e − →Ag(s) 0.80 C r 3+ (aq)+ e − →C r 2+ (aq) − 0.50 F e 3+ (aq)+3 e − →F e 2+ (aq) 0.77 C r 3+ (aq)+3 e − →Cr(s) − 0.73 C u +...
Using the following standard reduction potentials Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Pb2+(aq) + 2 e- → Pb(s) E° = -0.13 V calculate the standard cell potential for the galvanic cell reaction given below, and determine whether or not this reaction is spontaneous under standard conditions. Pb2+(aq) + 2 Fe2+(aq) → 2 Fe3+(aq) + Pb(s) Group of answer choices E° = -0.90 V, spontaneous E° = -0.90 V, nonspontaneous E° = +0.90 V, nonspontaneous E° = +0.90...
Using the following standard reduction potentials: Fe3+ (aq) + e. → Fe2+ (aq) Eo = +0.77 V Pb2+ (aq) + 2 e. → Pb(s) E。--0.13 V Calculate the standard cell potential for the galvanie cell reaction given below, and determine whether or not this reaction is spostaneous under standard conditions. Pb2+ (aq) + 2 Fe2+ (aq) → 2 Fe3+ (aq) + Pb(s) ⓔ A. E.-0.90 V, nonspontaneous OB. E-0.90 V, spontaneous C. Eo +0.90 V, nonspontaneous OD0.90 V, spontaneous
An electrochemical cell is based on these two half-reactions: Ox: Pb(s) Pb?" (aq, 0.21 mol L ')+2 e Erode = -0.13 (V) Red: Mno. (aq, 1.40 mol L-)+4 H+ (aq, 1.5 mol L-') +3 e MnO2 (s) + 2 H2O (1) Ecathode = 1.68 (V) Part A Compute the cell potential at 25°C. Express your answer to two decimal places and include the appropriate units. Ecell = Value Units Submit Request Answer
e) 20 g 4. Referring to the table of Standard Cell Potentials below, determine the standard cell potential for the following reaction: Pb2+(aq) + 2 Cl(aq) ---> Pb(s) + Cl2 E° -0.126 +1.360 Pb2+ (aq) + 2e ---> Pb (s) Cl2 + 2e- ---> 2 cl-(aq) a) +1.486 b) +1.468 c) +1.234 d) -1.234 e) -1.486
6. Consider the following galvanic cell and standard reduction potentials: Ag Pb E° = 0.80 V salt bridge Ag+ (aq) + e → Ag(s) Pb2+(aq) + 2e → Pb(s) E° = -0.13 V 1 M Ag+ 1 M Pb2+ Which one of the following statements is TRUE? a) The cell on the left containing Ag+(aq) is the anode. b) The initial reading on the voltmeter would be 0.67 V. c) Oxidation occurs in the cell on the right containing Pb²+(aq)....