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need answers for all questions Question 35 (3 points) (35) For galvanic cell: (-) Cr/ Cr3+...
please answer all wueston Question 33 (3 points) (33) For a galvanic cell notation: (-) Ni /Ni2+ (aq) // Au3+ (aq) / Au (+), The electrode potentials: Eº (Ni2+/ Ni) = -0.257 V, E ° (AU3+ / Au ) = 1.498 V. The cell potential Eºcell (a) 1.241 V (b) - 1.755 V O (0) 1.755 V (d) 1.498 Why Question 35 (3 points) (35) For galvanic cell: (-) Cr/ Cr3+ (aq) / Cu2+ (aq) / Cu(+), the correct half...
can you answer questions 7,8,9,10 Question 7 (10 points) ✓ Saved (7) For a galvanic cell notation: (-) Mg/Mg2+ (aq) // Cu2+ (aq) / Cu (+). which of the following is fully correct? (a) Mg2+ is oxidized to Mg, and Cu2+ (aq) is reduced to Cu O (b) Mg2+ is reduced to Mg, and Cu2+ (aq) is oxidized to Cu (c) Mg is oxidized to Mg2+, and Cu²+ (aq) is reduced to Cu (d) Mg is reduced to Mg2+, and...
HQ18.28 Homework • Unanswered Some electrode combinations that the Phoenix spacecraft designers may have considered for use in their design are given in the table below. Assume the ambient Martian temperature is -67.5*C, 205.7 K. What electrode combination (assume 0.1 M solutions) would provide the 1.317 V required by the spacecraft electronics? Use Appendix 11 for your standard cell potentials. А Rh+ +e- Rh/Ni2+ + 2e + NI O B Rh+ + e + Rh/Cr3+ + 3e - Cr C...
answer plz Question 32 (3 points) (32) For a galvanic cell notation: (-) Zn /Zn2+ (aq) // Cu2+ (aq) / Cu (+), which of the following is fully correct? (a) Zn2+ is oxidized to Zn, and Cu2+ (aq) is reduced to Cu. (b) Zn is oxidized to Zn2+, and Cu2+ (aq) is reduced to Cu. 0 (c) Zn2+ is reduced to Zn, and Cu2+ (aq) is oxidized to Cu. (d) Zn is reduced to Zn2+, and Cu (aq) is reduced...
Half-cell Potentials: Half Reaction: E value +0.80 V Agt + e → Ag Fe3+ + € → Fe2+ +0.77 v +0.34 V -0.13 V Cu2+ +2e → Cu Pb2+ + 2e - → Ib Ni2+ + 2e → Ni Cd2+ +2e → Cd -0.25 V -0.40 V Fe2+ + 2e → Fe -0.44 V Zn2+ + 2e → Zn -0.76 V Al3+ +3e → AI - 1.66 V Consider an electrochemical cell constructed from the following half cells, linked by...
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g) Express your answer using two significant figures. 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 Cu2+(aq)+2e−→Cu(s) 0.34 Mn2+(aq)+2e−→Mn(s) −1.18 2H+(aq)+2e−→H2(g) 0.00 Al3+(aq)+3e−→Al(s) −1.66 Fe3+(aq)+3e−→Fe(s) −0.036 Mg2+(aq)+2e−→Mg(s) −2.37 Pb2+(aq)+2e−→Pb(s) −0.13 Na+(aq)+e−→Na(s) −2.71 Sn2+(aq)+2e−→Sn(s) −0.14 Ca2+(aq)+2e−→Ca(s) −2.76 Ni2+(aq)+2e−→Ni(s) −0.23 Ba2+(aq)+2e−→Ba(s) −2.90 Co2+(aq)+2e−→Co(s) −0.28 K+(aq)+e−→K(s) −2.92 Cd2+(aq)+2e−→Cd(s)...
need help for half cell potentials pls calculate step by step (NOTE - Remember that the positive electrode is attached to the red wire and the negative electrode is attached to the black wire.) Electrode Systems Used Anode (oxidation) Negative Cathode (reduction) Positive Measured Potential (V) Positive (Ecu) Copper and silver Cu() - Cu2+ + 2e" Ag+ +1e Ag) 0.432 V Zinc + Silver Zn cs + 2n**+ Zé dat + leº nAg (s) 1.484 V Copper & Zinc Zn(s)...
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...
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction + O2(9)+4 H (aq)+4e' 2H20) = 1.23 V red Ered Fe+. (аq) Fe3(aq)+e = +0.771 V Answer the following questions about thiss cell Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous...
Consider a galvanic electrochemical cell constructed using Cr/Cr3* and Zn/Zn2+ at 25 °C. The following half-reactions are provided for each metal: Cr3+ (aq) + 3 e → Cr(s) Eºred = -0.744 V Zna*(aq) + 2 e Zn(s) Eºred = -0.763 V What is the standard cell potential for this cell?