Two standard reduction potentials are given below.
Pb2+(aq) + 2 e− → Pb(s) |
E⁰red = −0.126 V |
Cl2(g) + 2 e− → 2 Cl−(aq) |
E⁰red = +1.358 V |
(a) Which is a stronger reducing agent, Pb(s) or Cl−(aq)?
Pb(s) ; or Cl−(aq)
(b) Which is the most difficult to oxidize, Pb(s) or
Cl−(aq)?
Pb(s); or Cl−(aq)
(c) Is Pb(s) able to reduce Cl2(g) in
a spontaneous reaction?
is able; or is not able
(d) Is Cl−(aq)
able to reduce Pb2+(aq) in a spontaneous
reaction?
is able; or is not able
Two standard reduction potentials are given below. Pb2+(aq) + 2 e− → Pb(s) E⁰red = −0.126...
Two standard reduction potentials are given below. Cd2+(aq) + 2 e− → Cd(s) E⁰red = −0.403 V Al3+(aq) + 3 e− → Al(s) E⁰red = −1.662 V There is only ONE submission for each part. (a) Which is a stronger reducing agent, Cd(s) or Al(s)? Cd(s) Al(s) (b) Which is the most difficult to oxidize, Cd(s) or Al(s)? Cd(s) Al(s) (c) Is Cd(s) able to reduce Al3+(aq) in a spontaneous reaction? is able is not able (d) Is Al(s) able...
1. Consider the following standard reduction potentials, Pb 2 +(aq) + 2 e- → Pb(s) E° = -0.13 V I2(s) + 2 e- → 2 I-(aq) E° = +0.54 V Under standard conditions, Consider the following standard reduction potentials, Pb 2 +(aq) + 2 e- → Pb(s) E° = -0.13 V I2(s) + 2 e- → 2 I-(aq) E° = +0.54 V Under standard conditions, I2(s) is a stronger oxidizing agent than Pb 2+(aq) and Pb(s) is a stronger reducing...
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...
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
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
Using the standard reduction potentials below, Hg22+(aq) + 2 e- → 2 Hg() E° = +0.789 V I2(s) + 2 e- → 2 I-(aq) E° = +0.535 V Ni2+(aq) + 2 e- → Ni(s) E° = -0.25 V (a) which element or ion is the best oxidizing agent? __________ (b) Which element or ion is the best reducing agent? __________ (c) Which element or ion will oxidize I- ions? __________
4. Consider the following standard reduction potentials, Al3 (aq)+3 e Al(s) 12(s)2 e2 1(aq) E = -1.66 V +0.54 V iT E Under standard conditions: A. Al3(aq) is a stronger oxidizing agent than 12(s), and I(aq) is a stronger reducing agent than Al(s). B. 12(s) is a stronger oxidizing agent than Al3*(aq), and Al(s) is a stronger reducing agent than I'(aq). C. Al(s) is a stronger oxidizing agent than I-(aq), and Al3 (aq) is a stronger reducing agent than l2(s)....
Half-reaction E° (V) I2(s) + 2e- 2I-(aq) 0.535V Pb2+(aq) + 2e- Pb(s) -0.126V Cr3+(aq) + 3e- Cr(s) -0.740V The strongest oxidizing agent is: ______enter formula The weakest oxidizing agent is: The weakest reducing agent is: The strongest reducing agent is: Will I2(s) reduce Cr3+(aq) to Cr(s)? Which species can be reduced by Pb(s)? If none, leave box blank.
Pb2+(aq) + 2e− ⇌ Pb(s) E° = -0.126 V 2H+(aq) + 2e− ⇌ H2(g) E° = 0.000 V E°cell (in V)= 0.126 V 2. The electrochemical cell is comprised of a Pb electrode in a 1.67 × 100 M solution of Pb2+ (aq) coupled to a Pt electrode in a solution containing H+ (aq) where the pH of the solution is 0.37 and the partial pressure of H2(g) is 0.571 atm. The temperature of the cell is held constant at...
Calculate e cell for the electrochemical cell below, Pb(s) |Pb2+(aq, 1.0 M) || Fe2+(aq, 1.0 M) | Fe(s) given the following reduction half-reactions. Pb2+(aq) + 2 e– ® Pb(s) E° = –0.126 V Fe2+(aq) + e– ® Fe(s) E° = –0.44 V