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Any problem from number 5. Please explain step by step. I
don't understand steps. Thank you....
Fill in the Blanks 0.34 Half Reaction E. (V) Half Reaction Erd® (V) F + 2e...
Fill in the Blanks 0.34 Half Reaction E. (V) Half Reaction Erd® (V) F + 2e →2F 0+ 2H + 2e →H,02 0.68 Ag* + e → Ag Cu* + e → Cu 0.52 Co? + e Co? O2 + 2H,0 + 4e → 40H 0.40 H,O, + 2H + 2e → 2H,0 Cu2+ + 2e → Cu Cet+e → Ces Cu2+ + e → Cut 0.16 PbO, + 4H+ +50,2 +2e → PbSO, + 2H,0 2H* + 2e →...
1. How do I read the half reaction table? 2. If im asked for the best...
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...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
For all of the following experiments, under standard conditions, which species could be spontaneously produced? A...
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...
the first picture is about some useful information, and the second picture is the question that...
the first picture is about
some useful information, and the second picture is the question
that bothers me. I wonder how we know the half-cell reaction of it.
Please explain!!!
TABLE 18.1 | Standard Reduction Potentials at 25°C (298 K) for Many Common Half-Reactions 8° (V) 0.40 0.34 0.27 0.22 0.20 0.16 0.00 Half-Reaction F2 + 2e →2F Ag2+ + e +Agt Co3- + e + CO2- H2O2 + 2H+ + 2e +2H20 Ce+ + e + Ce+ PbO2 +...
2. Using the decay chain for 238U, calculate the amount of time it takes 5.0 kg...
2. Using the decay chain for 238U, calculate the amount of time it takes 5.0 kg of 238U to decay to 2.5 kg. What mass of 206Pb is produced? (5 points) (V) Helpful Stuff Thermodynamics: AG° = AH-TAS Nernst Equation: 6 = 6 - (RT/nF)InQ AG=RTIK At 25°C: 8 = 6 - (0.0591/n)logQ ΔGIRT Ke=e AGⓇ = -nF8° Units/Constants: Volt: 1 V=1/C Faraday: 1 F -96,485 C/mol e Arrhenius Equation: k = Ae Ea/RT R= 8.314 J/mol K Integrated Rate...
5. How much faster would a reaction be if a catalyst is used that lowers the...
5. How much faster would a reaction be if a catalyst is used that lowers the activation energy from 20.0 kJ/mol to 10.0 kJ/mol? Do the calculation at two temperatures: first at 25.0°C and then at 0.0°C. (20 points) (V) Helpful Stuff Thermodynamics: AG° = AH-TAS Nernst Equation: 6 = 6 - (RT/nF)InQ AG=RTIK At 25°C: 8 = 6 - (0.0591/n)logQ AGRT Ke=e AGⓇ =-nF8° Units/Constants: Volt: 1 V = 1 J/C Faraday: 1 F = 96,485 C/mole Arrhenius Equation:...
Find the best reducing agent from Cu+, Ag+ F2 and Fe3+ #1. In the reduction table...
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,...
Fill in the Blanks Ered (V) 0.68 0.52 0.40 0.34 0.16 0 Half Reaction Ered (V)...
Fill in the Blanks Ered (V) 0.68 0.52 0.40 0.34 0.16 0 Half Reaction Ered (V) Half Reaction F2+2e →2F 2.87 02 + 2H+ 2e →H,O, Ag?+ e → Ag 1.99 Cute → Cu Code → CO2 1.82 O2 + 2H,0 + 4e → 40H H,02 + 2H+ + 2e → 2H,0 1.78 Cu2+ + 2e → Cu Ce+ + → Ce+ 1.70 Cu2+ e → Cu PbO, + 4H* +502 +2e → PbSO4 + 2H,0 1.69 2H*+2e → H2...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery that... Reduction Half-Reaction F2(g) + 2e →2F(aq) S2082 (aq) + 2e- → 25042 (aq) O2(g) + 4H+ (aq) + 4e → 2H2O(1) Br2(1) + 2e + 2Br (aq) Agt(aq) + e → Ag(s) Fe3+ (aq) + e- → Fe2+ (aq) 126) + 2e → 21 (aq) Cu2+ (aq) + 2e → Cu(s) Sn4+ (aq) + 2e → Sn2+ (aq) S(s) + 2H+ (aq) +...
Fill in the Blanks 0.34 Half Reaction E. (V) Half Reaction Erd® (V) F + 2e →2F 0+ 2H + 2e →H,02 0.68 Ag* + e → Ag Cu* + e → Cu 0.52 Co? + e Co? O2 + 2H,0 + 4e → 40H 0.40 H,O, + 2H + 2e → 2H,0 Cu2+ + 2e → Cu Cet+e → Ces Cu2+ + e → Cut 0.16 PbO, + 4H+ +50,2 +2e → PbSO, + 2H,0 2H* + 2e →...
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...
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...
the first picture is about
some useful information, and the second picture is the question
that bothers me. I wonder how we know the half-cell reaction of it.
Please explain!!!
TABLE 18.1 | Standard Reduction Potentials at 25°C (298 K) for Many Common Half-Reactions 8° (V) 0.40 0.34 0.27 0.22 0.20 0.16 0.00 Half-Reaction F2 + 2e →2F Ag2+ + e +Agt Co3- + e + CO2- H2O2 + 2H+ + 2e +2H20 Ce+ + e + Ce+ PbO2 +...
2. Using the decay chain for 238U, calculate the amount of time it takes 5.0 kg of 238U to decay to 2.5 kg. What mass of 206Pb is produced? (5 points) (V) Helpful Stuff Thermodynamics: AG° = AH-TAS Nernst Equation: 6 = 6 - (RT/nF)InQ AG=RTIK At 25°C: 8 = 6 - (0.0591/n)logQ ΔGIRT Ke=e AGⓇ = -nF8° Units/Constants: Volt: 1 V=1/C Faraday: 1 F -96,485 C/mol e Arrhenius Equation: k = Ae Ea/RT R= 8.314 J/mol K Integrated Rate...
5. How much faster would a reaction be if a catalyst is used that lowers the activation energy from 20.0 kJ/mol to 10.0 kJ/mol? Do the calculation at two temperatures: first at 25.0°C and then at 0.0°C. (20 points) (V) Helpful Stuff Thermodynamics: AG° = AH-TAS Nernst Equation: 6 = 6 - (RT/nF)InQ AG=RTIK At 25°C: 8 = 6 - (0.0591/n)logQ AGRT Ke=e AGⓇ =-nF8° Units/Constants: Volt: 1 V = 1 J/C Faraday: 1 F = 96,485 C/mole Arrhenius Equation:...
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,...
Fill in the Blanks Ered (V) 0.68 0.52 0.40 0.34 0.16 0 Half Reaction Ered (V) Half Reaction F2+2e →2F 2.87 02 + 2H+ 2e →H,O, Ag?+ e → Ag 1.99 Cute → Cu Code → CO2 1.82 O2 + 2H,0 + 4e → 40H H,02 + 2H+ + 2e → 2H,0 1.78 Cu2+ + 2e → Cu Ce+ + → Ce+ 1.70 Cu2+ e → Cu PbO, + 4H* +502 +2e → PbSO4 + 2H,0 1.69 2H*+2e → H2...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery that... Reduction Half-Reaction F2(g) + 2e →2F(aq) S2082 (aq) + 2e- → 25042 (aq) O2(g) + 4H+ (aq) + 4e → 2H2O(1) Br2(1) + 2e + 2Br (aq) Agt(aq) + e → Ag(s) Fe3+ (aq) + e- → Fe2+ (aq) 126) + 2e → 21 (aq) Cu2+ (aq) + 2e → Cu(s) Sn4+ (aq) + 2e → Sn2+ (aq) S(s) + 2H+ (aq) +...
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