Half-reaction | E°(volts) |
U4+ + e- ---> U3+ | -0.607 |
U3+ + 3e- --->U | -1.798 |
U4+ + 4e- --->U | -1.500 |
In a disproportionation reaction, the same species is oxidized and reduced. Use the data given above to analyze the disproportionation reaction:
4U3+ --->U + 3U4+ |
Will U3+ disproportionate in aqueous solution? To justify your answer, what is the standard cell voltage for the disproportionation reaction?____ V
Standard Reduction Potentials at 25 c Half-reaction E°(volts) U4+ + e- ---> U3+ -0.607 U3+ + 3e-...
A Cacat Pb2+ Pb galvanic cell is constructed in which the standard cell voltage is 0.277 V Calculate the free energy change at 25°C when 0.866 g of Pb plates out, if all concentrations remain at their standard value of 1 M throughout the process. What is the maximum amount of work that could be done by the cell on its surroundings during this experiment? AGⓇE Maximum work Use the References to access important values if needed for this question....
In a galvanic cell, one half-cell consists of a lead strip dipped into a 1.00 M solution of Pb(NO). In the second half-cell, solid germanium is in contact with a 1.00 M solution of Ge(NO), Ge is observed to plate out as the galvanic cell operates, and the initial cell voltage is measured to be 0.250 V at 25°C. (a) Write balanced equations for the half-reactions at the anode and the cathode. Show electrons as e'. Use the smallest integer...
consider the following standard reduction potentials. Reduction Half-Reaction Eo (volts) Al3+(aq) + 3e− → Al(s) − 1.66 Fe2+(aq) + 2e− → Fe(s) − 0.44 Sn2+(aq) + 2e− → Sn(s) − 0.14 The Al/Al3+ half-reaction can be paired with the other two to produce voltaic cells because ________ A) Al is a more powerful oxidizing agent B) Fe and Sn are readily oxidized Al is a more powerful reducing agent C) Al3+ is a more powerful oxidizing agent D) Al3+...
pls explain in great detail why Consider the following standard reduction potentials Reduction Half-Reaction A1+ (aq) + 3e A1(s) Fe2+ (aq) + 2e-Fe(s) Sn2+ (aq) + 2e - Sn(s) E (volts) -1.66 0.44 0.14 The AIAP half-reaction can be paired with the other two to produce voltaic cells because Als* is a more powerful reducing agent OOO Al is a more powerful oxidizing agent o Fe and Sn are readily oxidized Al is a more powerful reducing agent AB+ is...
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)?
Standard Electrode Potentials at 25?C Reduction Half-Reaction E?(V) F2(g)+2e? ?2F?(aq) 2.87 Au3+(aq)+3e? ?Au(s) 1.50 Cl2(g)+2e? ?2Cl?(aq) 1.36 O2(g)+4H+(aq)+4e? ?2H2O(l) 1.23 Br2(l)+2e? ?2Br?(aq) 1.09 NO3?(aq)+4H+(aq)+3e? ?NO(g)+2H2O(l) 0.96 Ag+(aq)+e? ?Ag(s) 0.80 I2(s)+2e? ?2I?(aq) 0.54 Cu2+(aq)+2e? ?Cu(s) 0.16 2H+(aq)+2e? ?H2(g) 0 Cr3+(aq)+3e? ?Cr(s) -0.73 2H2O(l)+2e? ?H2(g)+2OH?(aq) -0.83 Mn2+(aq)+2e? ?Mn(s) -1.18 How can the table be used to predict whether or not a metal will dissolve in HCl? In HNO3? Drag the terms on the left to the appropriate blanks on the right to...
Given the following standard reduction potentials: Tl(3+) + e(-) ---> Tl(2+) E°= -0.37 V Tl(3+) + 2e(-) ---> Tl(+) E°= 1.25 V (a) Calculate the half-cell potential for the half-reaction Tl(2+) + e(-) ---> Tl(+) (b) Determine of the following disproportionate reaction will occur sponta- neously in aqueous solution in the standard condition. 2Tl(2+)(aq) ---> Tl(3+)(aq) + Tl(+)(aq)
4. A galvanic cell is composed of these two half-cells, with the standard reduction potentials shown: Cu2+(aq) + 2e Cu(s) Fe3+(aq) + 3e,-- Fe(s) +0.34V +0.77V What is the standard free energy change for the cell reaction of this galvanic cell?
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....
♡ Search this course Faded 0 x cell potentials given below. Standard Reduction Potentials (Volts) at 25 °C 12() + 2 6 2 I'(aq) 0.535 02(g) + 4 Hz0+ (aq) + 4 € -6 H,00 1.229 2 H2O(l) + 2 € H2(g) + 2 OH"(aq) -0.828 Sc3(aq) + 3 e Sc(s) -2.080 Half-reaction at anode: Half-reaction at cathode: Se*(aq) + 3 6 —- Sc(S) -2.080|| Half-reaction at anode: Half-reaction at cathode: (b) What is the expected decomposition potential?