Given the following pair of half-reactions, fill in the line
notation for a standard cell using these two half-reactions.
Since it is a standard cell, all concentrations will be 1 M, so do
not enter concentrations.
You have solid platinum (Pt) electrodes to use, where
necessary.
Do not include the physical state of the
components, i.e., solid platinum would just be Pt.
For subscript and superscripts use the model: Ag+ would
be Ag^+ and SO42- would be entered as SO4^2-,
with the "^" character indicating a superscript.
Leave a space between components.
Tl+ (aq) + e- → Tl (s) | ξo = -0.34 | |
Au3+ (aq) + 3e- → Au (s) | ξo = 1.50 |
Given the following pair of half-reactions, fill in the line notation for a standard cell using...
Course Contents > Electrochem IV-Graded HomERO Given the following pair of half-reactions, fill in the line notation for a standard cell using these two half-reactions Since it is a standard cell, all concentrations will be 1 M, so do not enter concentrations You have solid platinum (Pt) electrodes to use, where necessary Do not include the physical state of the components, i.e., solid platinum would just be Pt For subscript and superscripts use the model: Ag" would be Ag+ and...
Detailed answer please. 6. Suppose you initially have a standard galvanic cell based on the following half- reactions: Cu" (aq) 2eCu (s) Ag' (aq)e Ag (s) The metal electrodes in this cell are Ag (s) and Cu (s). Does the cell potential increase, decrease, or remain the same when the following changes occur? Provide a reasonable explanation to support your answers a) Solid CuSO4 is added to the copper half-cell compartment (CuSO4 b) NHs (aq) is added to the copper...
Use the standard half-cell potentials listed below to determine which the following metals will dissolve in hydrochloric acid. Cl2(g) + 2e + 2C1-(aq); E° = 1.36 V 2H+(aq) + 2e + H2(g); E° = 0.00 V O Cu; E°(Cu2+/Cu) = +0.34V O Au; E°(Au3+/Au) = +1.50V O Al; E°(A13+/Al) = -1.66V O Ag; E°(Ag+/Ag) = +0.80V O Pt; E°(Pt2+/Pt) = +1.19V
Use the standard half-cell potentials listed below to determine which the following metals will dissolve in hydrochloric acid. Cl2(g) + 2e -- 2014(aq); E° = 1.36 V 2H+(aq) + 2e - H2(g); E° = 0.00 V OPt; E°(P+2+/Pt) = +1.19V O Ag; E°(Ag+/Ag) = +0.80V O Au; EⓇ(Au3+/Au) = +1.50V O Al; E(A13+/Al) = -1.66V O Cu; E(Cu2+/Cu) = +0.34V
Use the following pair of reduction half-reactions to design a galvanic cell. Then write in the proper coefficient for each of the species involved in the overall reaction. Water molecules and protons are not shown in the half-reactions, but may be needed in the overall reaction. N2 (g) → N2H5+ (aq) ξo = -0.23 V MnO4- (aq) → Mn2+ (aq) ξo = 1.51 V MnO4- (aq) N2 (g) Mn2+ (aq) N2H5+ (aq) H+ (aq) H2O (l)
Use the following pair of reduction half-reactions to design a galvanic cell. Then write in the proper coefficient for each of the species involved in the overall reaction. Water molecules and protons are not shown in the half-reactions, but may be needed in the overall reaction. MnO4- (aq) → MnO2 (s) ξo = 0.59 V NO3- (aq) → NO (g) ξo = 0.96 V NO3- (aq) MnO4- (aq) NO (g) MnO2 (s) H+ (aq) H2O (l)
Given the following half reactions what is the voltage of the galvanic cell that would result from their combination [maximum positive voltage and spontaneous reaction). Show work. E° (volts) -0.74 Cr3+ (aq) + 3e Cu2+ (aq) + 2e → → Cr (8) Cu (s) +0.34 And what is the overall reaction that results after the half reactions are combined to cancel all electrons. Show work. And fill in the following abbreviated cell to represent the actual galvanic cell from above.
Given the following two half reactions with their standard reduction potentials: Rh3+(aq) + 3e− → Rh(s) E° = 0.76 V Au+(aq) + e− → Au(s) E° = 1.69 V What is the cell potential (E, in Volts) for a cell at 298.15 K whose anode consists of [Rh3+] = 0.40 M and Rh(s), and whose cathode consists of [Au+] = 0.10 M and Au(s)?
The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook: Ag+(aq)+e−→Ag(s)= .799 Cu2+(aq)+2e−→Cu(s)= .337 Ni2+(aq)+2e−→Ni(s)= -.28 Cr3+(aq)+3e−→Cr(s). = -.74 1. Determine which combination of these half-cell reactions leads to the cell reaction with the largest positive cell emf. 1st and 2nd, 1st and 3rd, 1st and 4th, 2nd and 3rd, 3rd and 4th. It isn't the first or last one because I have gotten it wrong twice. 2. Calculate the value of this emf....
Consider the following half-reactions and their standard reduction potentials then give the standard line (cell) notation for a voltaic cell built on these half reactions. Mn2+(aq) + 2 e- <=> Mn(s) E° = -1.18 V Fe3+(aq) + 3 e- <=> Fe(s) E° = -0.036 V Correct answer: Mn (s) | Mn 2+(aq, 1.0 M) || Fe3+(aq, 1.0 M) | Fe(s) looking for an explanation on how to work this problem, i get confused with the order of the elements. for...