What is the standard emf of a galvanic cell made of a Co electrode in a 1.0 M Co(NO32 solution and a Al electrode in a 1.0 M AI(NO3)3 solution at 25°C? 0 cell Standard Reduction Potentials at 25°C Half-Reaction E(V +2.87 +2.07 +1.82 O,(g) 2H (aq)2e0(g)+HO Co3+(aq) + e-_? Co2+(aq) H,02(aq) + 2H"(aq) + 2e-_ 2H20 Cu2+(aq) + 2e-? Cu(s) AgCIs) + Ag(s) + CI(a) S02-(aq) + 4H'(aq) + 2e S02(g) + 2H20 Cu2+(aq) + e-_ Cu+(aq) Sn (aq) eSn2 (aq) +0.34 +0.22 +0.20 +0.15 +0.13 0.00 -0.13 -0.14 -0.25 -0.28 -0.31 -0.40 -0.44 -0.74 -0.76 -0.83 ?1.18 1.66 Pb2+(aq) + 2e-- Pb(5) Co2+(aq) + 2e-? Cos) Cd2+(aq) + 2e-? Cd(s) Zn2+(aq) + 2e-_? Zn(s) Mn2+(aq) 2e--+ Mn(s) Al(a)3eAKS)
What is the standard emf of a galvanic cell made of a Co electrode in a 1.0 M Co(NO32 solution and a Al electrode in a 1.0 M AI(NO3)3 solution at 25°C?
(12) Calculate the cell potential (Eci) at 25C ofa galvanic cell consisting of an Ag electrode in 0.15 M AgNOs and an Ag electrode in 1.0 MAgNOs. (7 points) Canvas | elearning M 12. CHM 122,01 Examos Sprinc × 1 + ok.do?setTab-sectionTabs jump to pg go book contents search obook go Half-Reaction +2.87 +2.07 +1.82 +1.70 +1.61 +1.50 +1.36 +1.33 +1.23 +1.23 +1.07 +0.96 +0.92 +0.85 +0.80 +0.77 Crot(aq) + 14HOaq) + 6e-一2Cr .daq) + 까1,0) 16)+2a +0.53 +0.22 +0.20...
In the following cell, A is a standard Cu2+Cu electrode connected to a standard hydrogen electrode. If the voltmeter reading is +0.34 V, which half-reaction occurs in the left-hand cell compartment? Given: Standard reduction potential of the H1/H2 and Cu2*/Cu couples are 0.00 and +0.34 V, respectively. Holo H2(g) --> 2H+ (aq) + 2e7 2H(aq) + 2e --> H2(g) Cu(s) --> Cu2(aq) + 2e Cu2(aq) + 2e --> Cu(s)
Use the galvanic cell notation to describe the Daniell cell. (4 marks) Using the reduction potential table calculate Eº for the Daniel cell. 12 marks) Calculate the equilibrium constant K for the reaction Cu2+ + Zn → Cu + Zn²+, at 25 °C. (4 marks) Standard Potentials at 25°C Half Reaction Potential Potential +2.87 V 0.000 V +2.07 V -0.04 V +2.05 V -0.13 V -1.69 V -0.14 V +1.69 V -0.23 V +1.67 V -0.26 V +1.63 v -0.28...
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...
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...
croHissit Song "14. a) Calculate the standard emf and write the overall equation for the cell described as: Croaq) Haq) + (aq) → Craq) + 2/8) + H2O(D Cr₂O7991+ 4H₂011 +36-7 reducing Oxidizing 2 croren + 4H20 (0) +36 --> CrotsstSoH + 5% 3(21 691 -> Izintze-) 0.406 14 7 .0 I BOV overall eqn: 200 2 06112 --> 2010 He(s) tel +3120) emf: 0.106 b) Calculate the emf obtained by this cell (based on part a above) from the...
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)...
Given the following standard reduction potentials choose the cell which will work as a voltaic cell. All cells below are written according to the usual cell diagram convention. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V 2H+(aq) + 2e → H2(g) E° = 0.00 V Sn2+ (aq) + 2e → Sn(s) E° = -0.14 V Ni2+(aq) + 2e → Ni(s) E° = -0.26 V Cd2+(aq) + 2e → → Cd(s) E° = -0.40 V Sn(s) | Sn2+(aq) || Ni2+(aq)...
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)...
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...