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Question 22 (5 points) Given the following half reactions: PbO2 + 4H+ + SO42- + 2e...
Question 20 (5 points) Using the following standard reduction potentials: PbO2 + 4H+ + SO42- + 2e → PbSO4(s) + 2H20 E° = +1.69 V PbO2 + 4H+ + 2e → Pb2+ + 2H20 E° = +1.46 V The standard cell voltage for the reaction PbSO4(s) Pb2+ + SO42- is V. Please enter your answer in decimal form, such as 2.5, and keep the correct number of significant figures.
Question 20 (5 points) Using the following standard reduction potentials: PbO2 + 4H+ + SO42- + 2e → PbSO4(s) + 2H20 E° = +1.69 V PbO2 + 4H+ + 2e Pb2+ + 2H20 E° = +1.46 V is The standard cell voltage for the reaction PbSO4(s) → Pb2+ + SO42- V. Please enter your answer in decimal form, such as 2.5, and keep the correct number of significant figures.
Question 22 (5 points) Given the following half reactions: - AgBr(s) + e- Ag+ (aq) + e- Ag(s) + Br- (aq) E° = 0.07V → Ag (s) E° = 0.80V - Calculate the Ksp for PbSO4 at 25 degrees C. Please enter your answer in scientific notation, such as 1.2E10 or 1.2E-10 and keep two significant figures.
You want to make a battery using the following reactions: PbO2 + 4H+ + SO42- + 2e ---> PbSO4(s) + 2H20 Ni2+ + 2e ---> Ni ° = +1.69 E° = -0.25 Which is being reduced? Which is being oxidized? Which is the anode? Which is the cathode? What is the formula to calculate Eºcell? What is Eºcell?
5. A lead storage battery involves the following two half-reactions: PbSO4(s) + 2e → Pb(s) + SO42- (aq); E = -0.36 V PbO2(s) + 4H*(aq) + SO42 (aq) + 2e → PbSO4(s) + 2H2O(1); E° = 1.69 V In the lead battery during the discharge reaction: A) PbSO4 is the cathode. B) PbSO4 is the anode. C) Pb is the anode. D) PbO, is the anode. E) H2SO4 is the cathode.
For all of the following experiments, under standard conditions, which species could be spontaneously produced? A lead wire is placed in a solution containing Cu2+ yes no Cu yes no PbO2 yes no No reaction Crystals of I2 are added to a solution of NaCl. yes no I- yes no No reaction yes no Cl2 A silver wire is placed in a solution containing Cu2+ no yes Cu no yes No reaction no yes Ag+ Half-Reaction 8° (V) Half-Reaction 8° (V) 2.87 1.99 1.82 1.78 1.70 1.69 1.68 1.60...
1. How do I read the half reaction table? 2. If im asked for the best reducing agent from Cu+, Ag+, F2, and Fe3+, where do I look first in the table? before the arrow or after the arrow? 3. Sometimes a value that has originally a positive (V) from the table it will have the negative sign in a homework problem, and viceversa, so the question is, how do I use the positive and negative signs in respect to...
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.
Find the best reducing agent from Cu+, Ag+ F2 and Fe3+ #1. In the reduction table i can see several repeated values of Fe3+ one is equal to 0.77v and the second one is equal to -0.036v so, which one do I choose? Please explain. #2.If I'm asked to find the best oxidation agent, from the values already provided (Cu+, Ag+ F2 and Fe3+) which one would it be? and how would I decide from repeated values, like in #1,...
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.