oxidation reaction : 2Na -- -> 2Na+ + 2e ..... Eo1 = -2.71 V
reduction reaction : 2H2O + 2e -> H2 + 2OH- .....Eo2 = - 0.83 V
E cell = E cathode – E anode
= -0.83 –(-2.71)V
= 1.88 V
Now calculate the dG 0 as follows:
dG 0 = - nF E cell
= - 2 mole *96500 c/ mole *1.88 V
= - 362840 J
= -362.84 kj
This reaction is spontaneou, because it results in a decrease of gibbs free energy or negative
expected for a reaction in which OR PRACTICE 39.6 Use tabulated electrode potentials to calculate 4G...
Part A Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2Li(s)+2H2O(l)→H2(g)+2OH−(aq)+2Li+(aq) Express your answer to three significant figures and include the appropriate units. Part B Is the reaction spontaneous?
Use tabulated standard electrode potentials to calculate the standard cell potential for this reaction occurring in an electrochemical cell at 25 ∘C. Cr(s)+NO−3(aq)4H+(aq)→Cr3+(aq)+NO(g)+2H2O(l) Express your answer using two decimal places and include the appropriate units.
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...
1. Use the tabulated half-cell potentials to calculate AG for the following balanced redox reaction. SHow your work 2 Fe3+(aq) + 6 1(aq) 3 12(s)+ 2 Fe(s)
use tabulated standard electrode potential to calculate the
standard cell potential for the reaction occurring in an
electrochemical cell at 25 C. (The equation is balanced.)
3Ni^2+(aq)+2Cr(s)--->3Ni(s)2Cr^3+(aq)
Express your answer to two significant figures and include the
appropriate units.
em 26 E (V) -0.45 -0.50 -0.73 -0.76 -1.18 Standard reduction half-cell potentials at 25°C Half-reaction E° (V) Half-reaction Aul+ (aq) + 3e +Au(s) 1.50 Fe2+ (aq) + 2eFe(s) Ag+ (aq) +e-Ag(s) 0.80 Cr3+ (aq) + Cr²+ (aq) Fe+(aq) + 3e...
Use tabulated half-cell potentials (found in appendix IV-D) to calculate Eºcell and then calculate AGºrxn for the following reaction at 25 °C. 2Fe3+ (aq) + 3Sn(s) → 2Fe(s) + 3Sn2+ (aq) Hint: the standard potential of Fe3+ + 3e + Fe(s) is -0.036 and the standard potential of Sn2+ + 2e → Sn(s) is -0.140. AG° = -5 kJ/mol AG° = -10 kJ/mol AG° = -20 kJ/mol AG° = -30 kJ/mol
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)...
K for the following reaction: 7. Use the tabulated half-cell potentials to calculate 2 Cr(s) +3 Fe (aq) 2 Cr3(aq) 3 Fe(s)
Use the tabulated half-cell potentials to calculate AGº for the following balanced redox reaction 3 12(s) + 2 Fe(s) → 2 Fe3+(aq) + 6 H(aq) +23 x 102 kJ 0 +49 x 101 kJ -1.1 x 102 kJ 0 -9.7 x 101 kJ 0 -3.3 x 102 kJ
Using the following data (tabulated at 25°C), calculate AG, ΔΗ', and AS0 for each of the reactions listed in number 4. Record the answers in your lab notebook. You will need them for your report. Species Cuo 0.0 +65.5 24.4-147.1 153.9 +0.0105-0.1121 0.0 AHo J-mor0.0 +6481.7 - Se (J.K mo 0.0332 0.0648 0.0416 -0.0996 + 0.0101.6 u sehrences showing any calculations that are needed to 1. For each cell you will study today, identfy which electrode that which will serve...