We need at least 10 more requests to produce the answer.
0 / 10 have requested this problem solution
The more requests, the faster the answer.
Consider the following half-reactions and their standard reduction potentials then give the standard line (cel notation...
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...
pleaae do all, please please please QUESTION 3 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 Cr3+ (aq) + 3 e <-> Cris) - 0.41 V A Mn (s) | Mn 2+(aq, 1.0 M) 1 Cr3+(aq. 1.0 M) | Cris) B. Mn (s) Mn 2+(aq, 1.0 M) || Cris) | Cr3+(99, 1.0 M)...
Standard (reduction) potentials are given below for several half-reactions. Use them to answer the following. Mn2+ (aq) + 2e- → Mn(s) Cr3+(aq) + 3 e- → Cr(s) Sn2+(aq) + 2 e--> Sn(s) Cu2+ (aq) + 2 e → Cu(s) A13+ (aq) + 3 e- → Al(s) Co2+(aq) + 2 e- → Co(s) Cd2+ (aq) + 2 e- → Cd(s) Ered = -1.185 V Ered = -0.744 V Ered = -0.138 V = +0.342 V Ered = -1.662 V Ered =...
Standard (reduction) potentials are given below for several half-reactions. Use them to answer the following. Cd2+ (aq) + 2 e- → Cd(s) Mn2+(aq) + 2 e- → Mn(s) Cu2+ (aq) + 2 e- → Cu(s) Cr3+(aq) + 3 e- → Cr(s) Sn2+(aq) + 2 e- → Sn(s) A13+(aq) + 3 e- → Al(s) Co2+ (aq) + 2 e- → Co(s) Ered = -0.403 V Ered = -1.185 V Ered = +0.342 V Ered = -0.744 V Ered = -0.138 V...
Exercise 18.15 The voltaic cell is represented with the line notation. Standard Reduction Half-Cell Potentials at 25 °C Half Reaction HNO2 (aq) + 2 H+ (aq) + e--+ NO(g) + H20(1) NO. (aq) + 4 H + (aq) + e-? NO(g) + 2 H2O(1) sn't (aq) + 2 e-? Sn2 + (aq) 2H+ (aq) + 2e-?H2(g) Sn2+ (aq) + 2 e-? Sn(s) E (V) 0.98 0.96 0.15 0.00 -0.14
21) Which of these species is the strongest reducing agent? Given Standard Reduction Potentials Mg2+ (g)+2eMg (s) 2.37v Zn2+ (g) +2e Zn (s) 0.76 v Cr3 (aq) +3e Cr (s) E0.73 v Mn2+ (aq) + 3e-→ Mn (s) E。=-1.18 V A. Cr C. Mn D. Zn E. None of These
Use the standard reduction potentials from the following table to choose the weakest reducing agent among those shown below. Ag+(aq) + - + Ag(s) E° = 0.80 V Cu2+(aq) + 2e → Cu(s) E° = 0.34 V Ni2+(aq) + 2e + Ni(s) E° = -0.26 V Cr3+(aq) + 3e- → Cr(s) E° = -0.74 V Mn2+(aq) + 2e → Mn(s) E° = -1.19 V О Ni(s) Ag(s) Cr(s) Mn(s) Cu(s)
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) +...
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....
2 A voltaic cell is set up with one beaker containing 1.0 M Cu(NO 3) 2 and a copper electrode, and another beaker containing 1.0 M Mn(NO 3) and a manganese electrode. Given the following standard reduction potentials, answer the 3 questions below: E Cu2+(aq) + 2e Cu(s) 0.34 V Mn2+(aq) + 2e + Mn(s) 1.18V Part a. Write out the half-cell reaction that occurs at the anode of the voltaic cell. Part b. In which direction do electrons flow?...