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reduction potential of zinc is -.76
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Please help on these! thanks!
reduction potential of zinc is -.76
Cu cell V Cu V...
hurry!! The standard reduction potential of the Cu2+1Cu electrode is +0.34 V and the standard potential...
hurry!!
The standard reduction potential of the Cu2+1Cu electrode is +0.34 V and the standard potential of the cell Pb(s) Pb2+(aq) || Cu2+(aq) | Cu(s) is +0.47 V. What is the standard reduction potential of the Pb2+|Pb electrode? 0-0.81V +0.81 V O +0.13 V 0 -0.26 V O-0.13 V
The standard reduction potential of the Pb2+|Pb electrode is –0.13 V and the standard potential of...
The standard reduction potential of the Pb2+|Pb electrode is –0.13 V and the standard potential of the cell Zn(s) | Zn2+(aq) || Pb2+(aq) | Pb(s) is +0.63 V. What is the standard reduction potential of the Zn2+|Zn electrode? Question 9 options: –0.76 V –1.52 V +0.76 V –0.50 V +0.50 V
If a platinum indicator electrode is used to measure [Cu^22^++] (standard reduction potential of Cu^22^++ is 0.337 V), and [Cu^22^++] = 0.0053 M, what is the cell potential (V) if the reference electr...
If a platinum indicator electrode is used to measure [Cu^22^++]
(standard reduction potential of Cu^22^++ is 0.337 V), and
[Cu^22^++] = 0.0053 M, what is the cell potential (V) if the
reference electrode is SCE and the measurement is taken at 298
K?
Indicator Electrode--Quantification (Homework) Homework Unanswered If a platinum indicator electrode is used to measure [Cu2+] (standard reduction potential of Cu +is 0.337 V), and [Cu2+ o.0053 M, what is the cell potential (V) if the reference electrode...
Calculate the cell potential difference, Eºcell (Ecell = Eºcathode - Eanode), given the standard reduction potential...
Calculate the cell potential difference, Eºcell (Ecell = Eºcathode - Eanode), given the standard reduction potential below, Cut2 (aq) + 2 e. ----> Cu (s) ° = 0.337 volts Pb+2 (aq) + 2 e--...-> Pb(s) ° = -0.126 volts 0.211 V 0.463 V -0.463 V -0.211 V In Part III, to light up a white LED bulb will require forward voltage (VF) at 20 mA. • 5 1.2. 2.0 Calculate the cell potential difference, Eºcell ( cell = Eºcathode -...
HIU Blud cell potential of a galvanic cell based on the following reduction half- reactions at...
HIU Blud cell potential of a galvanic cell based on the following reduction half- reactions at 25 °C Cd + 2e → Cd Eº - -0.403 V Pb2+2e → Pb Eº - -0.126 V where (Cd-) - 0.040 M and (Pb) - 0.400M (13.) A zinc electrode is submerged in an acidic 0.40 M Zn?' solution which is connected by a salt bridge to a 1.50 M Ag' solution containing a silver electrode. Determine the initial voltage of the cell...
Table 1 Cell Type Lithium-iodine Zinc-mercury Operating Cell Potential for Commercial Batteries, E (V) +2.80 +1.35...
Table 1 Cell Type Lithium-iodine Zinc-mercury Operating Cell Potential for Commercial Batteries, E (V) +2.80 +1.35 Table 2 Standard Reduction Potential, E" (V) Half-Reaction Zn(OH).1? +2e → Zn +40H Zn(OH), +2e → Zn +20H HgO + H2O +2e - Hg + 2OH 0, + 2 H2O + 4e 40H - 1.20 |--1.25 +0.10 +0.40 Pacemakers are electronic devices that help regulate the heart rate. Currently, lithium-iodine cells are commonly used to power pacemakers and have replaced zinc-mercury cells. Table 1...
Please help, thanks. Use the table below to answer the questions that follow. Half Reaction &
Please help, thanks. Use the table below to answer the questions that follow. Half Reaction E°(V) Ag+ + e- → Ag (s) +0.80 Cu2+ + 2e- → Cu (s) +0.34 Pb2+ + 2e- → Pb (s) -0.13 Ni2+ + 2e- → Ni (s) -0.28 (A) Calculate E°cell for a voltaic cell made with Ni, Ni2+and Ag, and Ag+. (B) Which substance is Oxidized? (be sure to include the state of the oxidized substance in your answer) Zn(s) + CuSO4(aq) -->...
Chemistry: Experimental Cell Potential Lab report please help checking my work and answering question #2 #3 #4 Thanks!...
Chemistry:
Experimental Cell Potential Lab report
please help checking my work and answering question #2 #3
#4
Thanks!
*question #2 #3 #4*
Note that the ionic form of Ag is Ag' and of Fe is Fe Write a chemical reaction for each cell. For the reactants, choose the metal that was oxidized ion that was reduced spontaneously (i.e, a + potential) according to your data. (Thesethe elements!) Chemical reaction (oall) Cell notation (see text) (NA1 Zn(s)+Cu-Zn+Cu(s) r凸の, Zn(s) | Zn2...
help please A voltaic cell is constructed based on the oxidation of zinc metal and the...
help please
A voltaic cell is constructed based on the oxidation of zinc metal and the reduction of silver cations. Solutions of silver nitrate and zinc nitrate also were used. Locate the zinc nitrate on the diagram. c. c
10. Consider the standard potential data shown below. va* + 2e-Ⓡ v E' =-1.18 V Cu...
10. Consider the standard potential data shown below. va* + 2e-Ⓡ v E' =-1.18 V Cu + le ® Cut E* = 0.15 V -E° E Ech Use the above data to determine the standard emf for the reaction shown below. Is the reaction Spontaneous? V + 2cuva + 20 (a) Ex--1.03 V therefore the reaction is not spontaneous. (b) Ed. = 1.03 V therefore the reaction is spontaneous. = 1.33 V therefore the reaction is spontaneous. (d) Ex=-1.33 V...
hurry!!
The standard reduction potential of the Cu2+1Cu electrode is +0.34 V and the standard potential of the cell Pb(s) Pb2+(aq) || Cu2+(aq) | Cu(s) is +0.47 V. What is the standard reduction potential of the Pb2+|Pb electrode? 0-0.81V +0.81 V O +0.13 V 0 -0.26 V O-0.13 V
If a platinum indicator electrode is used to measure [Cu^22^++]
(standard reduction potential of Cu^22^++ is 0.337 V), and
[Cu^22^++] = 0.0053 M, what is the cell potential (V) if the
reference electrode is SCE and the measurement is taken at 298
K?
Indicator Electrode--Quantification (Homework) Homework Unanswered If a platinum indicator electrode is used to measure [Cu2+] (standard reduction potential of Cu +is 0.337 V), and [Cu2+ o.0053 M, what is the cell potential (V) if the reference electrode...
Calculate the cell potential difference, Eºcell (Ecell = Eºcathode - Eanode), given the standard reduction potential below, Cut2 (aq) + 2 e. ----> Cu (s) ° = 0.337 volts Pb+2 (aq) + 2 e--...-> Pb(s) ° = -0.126 volts 0.211 V 0.463 V -0.463 V -0.211 V In Part III, to light up a white LED bulb will require forward voltage (VF) at 20 mA. • 5 1.2. 2.0 Calculate the cell potential difference, Eºcell ( cell = Eºcathode -...
HIU Blud cell potential of a galvanic cell based on the following reduction half- reactions at 25 °C Cd + 2e → Cd Eº - -0.403 V Pb2+2e → Pb Eº - -0.126 V where (Cd-) - 0.040 M and (Pb) - 0.400M (13.) A zinc electrode is submerged in an acidic 0.40 M Zn?' solution which is connected by a salt bridge to a 1.50 M Ag' solution containing a silver electrode. Determine the initial voltage of the cell...
Table 1 Cell Type Lithium-iodine Zinc-mercury Operating Cell Potential for Commercial Batteries, E (V) +2.80 +1.35 Table 2 Standard Reduction Potential, E" (V) Half-Reaction Zn(OH).1? +2e → Zn +40H Zn(OH), +2e → Zn +20H HgO + H2O +2e - Hg + 2OH 0, + 2 H2O + 4e 40H - 1.20 |--1.25 +0.10 +0.40 Pacemakers are electronic devices that help regulate the heart rate. Currently, lithium-iodine cells are commonly used to power pacemakers and have replaced zinc-mercury cells. Table 1...
Chemistry:
Experimental Cell Potential Lab report
please help checking my work and answering question #2 #3
#4
Thanks!
*question #2 #3 #4*
Note that the ionic form of Ag is Ag' and of Fe is Fe Write a chemical reaction for each cell. For the reactants, choose the metal that was oxidized ion that was reduced spontaneously (i.e, a + potential) according to your data. (Thesethe elements!) Chemical reaction (oall) Cell notation (see text) (NA1 Zn(s)+Cu-Zn+Cu(s) r凸の, Zn(s) | Zn2...
help please
A voltaic cell is constructed based on the oxidation of zinc metal and the reduction of silver cations. Solutions of silver nitrate and zinc nitrate also were used. Locate the zinc nitrate on the diagram. c. c
10. Consider the standard potential data shown below. va* + 2e-Ⓡ v E' =-1.18 V Cu + le ® Cut E* = 0.15 V -E° E Ech Use the above data to determine the standard emf for the reaction shown below. Is the reaction Spontaneous? V + 2cuva + 20 (a) Ex--1.03 V therefore the reaction is not spontaneous. (b) Ed. = 1.03 V therefore the reaction is spontaneous. = 1.33 V therefore the reaction is spontaneous. (d) Ex=-1.33 V...
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