Given the following electrode potentials at 25°C
Fe3+ + e- → Fe2+ E° = 0.171 V
Fe2+ + 2e-→ Fe(s) E° = -0.440 V
Calculate the electrode potential for
Fe3+ + 3 e- → Fe(s)
Select one:
a.-0.236
b. 0.081
c.-0.036
d. -0.211
e. 0.121
+ Given the following electrode potentials at 25°C Fe3+ e-- Fe2+ E° = 0.571 V 2e Fe(s) E° = -0.440 V Calculate the electrode potential for Fe3+ + 3 e- Fe(s) Fe2+ + Select one: a. -0.132 b. -0.036 c. 0.081 d.-0.211 e. 0.103
7.53 Given the following electrode potentials at 25°C, Fe3+ +e- = Fe2+ E' = 0.771 V Fe2+ + e = Fe(s) E = -0.440 V calculate the electrode potential for [Fe3+ + e = {Fe E; = ? for
Calculate the equilibrium constant for each of the reactions at 25∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Fe3+(aq)+3e− →Fe(s) -0.036 Sn2+(aq)+2e− →Sn(s) -0.14 Ni2+(aq)+2e− →Ni(s) -0.23 O2(g)+2H2O(l)+4e− →4OH−(aq) 0.40 Br2(l)+2e− →2Br− 1.09 I2(s)+2e− →2I− 0.54 A) 2Fe3+(aq)+3Sn(s)→2Fe(s)+3Sn2+(aq) (answers are not 4.1x10^5, 3.3x10^3, 2.7x10^10, or 2.6x10^10) B) O2(g)+2H2O(l)+2Ni(s)→4OH−(aq)+2Ni2+(aq) C) Br2(l)+2I−(aq)→2Br−(aq)+I2(s) (answer is not 1.7x10^18)
Calculate the equilibrium constant for each of the reactions at 25∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Fe3+(aq)+3e− →Fe(s) -0.036 Sn2+(aq)+2e− →Sn(s) -0.14 Cu2+(aq)+2e− →Cu(s) 0.16 O2(g)+2H2O(l)+4e− →4OH−(aq) 0.40 Cl2(g)+2e− →2Cl− 1.36 I2(s)+2e− →2I− 0.54 Part B O2(g)+2H2O(l)+2Cu(s)→4OH−(aq)+2Cu2+(aq) (Express your answer using two significant figures. ) Part C Cl2(g)+2I−(aq)→2Cl−(aq)+I2(s) (Express your answer using two significant figures.)
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...
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)...
Standard potentials are measured against the standard hydrogen electrode (SHE). Because it is not always convenient to use a S.Н.Е., often other reference electrodes are used. The saturated calomel electrode (SCE.) is one commonly used reference electrode, with a reduction potential of +0.242 V versus the S.H.E. Using a table of standard reductions, determine what the standard reduction potential of each reduction half-reaction would be versus the S.C.E Cl2 (g) + 2e-→ 2 Cl-(aq) E1.36 Fe3 + (aq) + 3...
Given the following list of half-reaction reduction potentials, identify the reaction that will occur spontaneously as written: Half-reaction E° (V) -0.74 Cr3+ (aq) + 3 e ---> Cr (s) Sn4+ (aq) + 2 e ---> Sn2+ (aq) +0.154 -0.440 Fe2+ (aq) + 2 e ---> Fe(s) Fe3+ (aq) + e ---> Fe2+ (aq) +0.771 2 Cr (s) + 3 Fe2+ (aq) ---> 3 Fe (s) + 2 Cr3+ (aq) 2 Cr3+ (aq) + 3 Sn2+ (aq) ---> 3 Sn4+ (aq)...
Using std. electrode potentials listed, calculate the equilibrium constant at 25 ⁰C for the following reaction: O2 (g) + 4 H+ (aq) + 4 Fe2+ (aq) → 4 Fe3+ (aq) + 2 H2O (l) Fe3+ + e- = Fe2+ Ered= 0.77 O2 + 4H+ + 4e- = 2 H20 Ered= 1.23
Standard reduction half-cell potentials at 25°C E (V) E (V) 1.50 -0.45 0.80 -0.50 0.77 -0.73 0.52 -0.76 0.34 -1.18 Half-reaction Aut (aq) + 3e +Au(s) Ag+ (aq) + +Ag(s) Fe3+ (aq) +34 Fo+ (aq) Cut(aq) + Cu(s) Cu²+ (aq) + 2e +Cu(s) 2H+ (aq) - 2e +H2 (6) Fe3+ (aq) + 3e Fe(s) Pb2+ (aq) + 2e →Pb(s) Sn-(aq) + 2e +Sn(s) Ni2+ (aq) + 2e →Ni(s) Co2(aq) +2e + Co(s) ca? (aq) + 2e +Cd(s) 0.00 Half-reaction Fe(aq)...