Complete each of the following half-reactions with the correct number of electrons and then arrange them...
Complete each of the following half-reactions with the correct number of electrons and then arrange them from strongest oxidizing potential to strongest reducing potential. Strongest Oxidizing Potential Drag the text blocks below into their correct order 5e Ag (aq) Ag(s) E=+0.80 V 2e Sn4 (aq) Sn2 (aq) E = +0.13 V Ca2 (aq) Зе Ca(s) E = -2.87 V Cr(s) E -0.74 V Cr3 (aq) 4e -2F (aq) E= +2.87 V F2(g) e » 21-(aq) E = +0.53 V 12(s)...
Complete each of the following half-reactions with the correct number of electrons and then arrange them from strongest oxidizing potential to strongest reducing potential. Strongest Oxidizing Potential Drag the text blocks below into their correct order. 5e 2H*(aq) H2(g) E +0.00 V бе Cu2 (aq) Cu (aq) E = +0.15 V 4e Au3 (aq) Au(s) E 1.50 V 2e (s) 21(aq) E +0.53 V e Fe3 (aq) Fe2 (aq) E° = +0.77 V 2F (aq) E = +2.87 V F2(g)...
Complete each of the following half-reactions with the correct number of electrons and then arrange them from strongest oxidizing potential to strongest reducing potential. Strongest Oxidizing Potential Drag the text blocks below into their correct order. 2e 12(s) + → 21 (aq) E° = +0.53 V 4e Fe (aq) + → Fez (aq) E° = +0.77 V F2(g) + 2F-(aq) E° = +2.87 V AHOROD WWWWWWWWWWWWWWWWWWWWWWWWWWWWWWY Au3+ (aq) + Au(s) E° = +1.50 V 2H'(aq) + - + H2(g) E°...
Complete each of the following half-reactions with the correct number of electrons and then arrange them from strongest caidizing potential to strongest reducing potential Strongest Oxidizing Potential Drag the text blocks below into their correct order. Fes (aq) + + Fe (aq) = +0.77 V 2H(aq) + * H(o) +0.00 V Natal - Nais) E.-2.71 V Co (aq) + -Co" (aq) +1.82 V Br(0) + - 2Br(ad E - +1.07 v Nad) - Nis) E--0.25 V Strongest Reducting Potential
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...
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+...
Question 30 1 pts Rank the following in order of strength as a reducing agent. E° (V) Ag (aq) +e-4Ag(s) +0.80 Al(aq) + 3e Al(s) -1.66 Au3(aq) + 3e-Au(s) +1.50 Co2(aq) + 2e- 4 Cols) -0.28 Cr2+ (aq) + 2e-4Cr(s) -0.91 Ni?(aq) + 2e-Ni(s) -0.25 Pt2(aq) + 2e-4Pt(s) +1.19 Sn2(aq) + 2e-4Sn(s) -0.14 Ag [Choose 2 1 - strongest 3 - weakest TRUSC AI Co [Choose]
Sodium metal is prepared by electrolysis of sodium chloride to the metal and chlorine gas. Using the reduction potentials attached, calculate E° and ΔG° for the overall reaction. TABLE 12.2 Standard Reduction Potentials at 25°C Half-reaction Eº, V Increasing strength as oxidizing agent Lit(aq) + e Li(s) K+(aq) + e → KS) Ba2+ (aq) + 2e → Ba(s) Sr2+(aq) + 2e → Sr(s) Ca2+(aq) + 2e → Ca(s) Na*(aq) + e Na(s) Mg2+ (aq) + 2e → Mg(s) Be2+(aq) + 2e...
Using standard reduction potential in aqueous solutions at 25c Table, which substance is most likely to be oxidised by O2 (g) in acidic aqueous solution? Select one: a. Br2 (l) b. Br- (aq) c. Ni2+ (aq) d. Ag (s) e. Cu2+ (aq) Cathode (Reduction) Half-Reaction Standard Potential E° (volts) Li+(aq) + e- -> Li(s) -3.04 K+(aq) + e- -> K(s) -2.92 Ca2+(aq) + 2e- -> Ca(s) -2.76 Na+(aq) + e- -> Na(s) -2.71 Mg2+(aq) + 2e- -> Mg(s) -2.38 Al3+(aq)...
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...