Given the following reduction potentials: HNO_3 + 3H^+ + 3e^- NO + H_2O 0.96 2H^+ H_2...
MnO_2(s) + 2H_2 O(I) +2e = Mn^2+ (aq) + 40H^- (aq) epsilon^o = 1.208 v 2HNO_3 (aq) + 2H^+ (aq) + 2e = N_2 O_4 (g) + 2H_2 O(I) epsilon^o = 0.96 v N_2 O_4(g) + 2H^+ (aq) + 2e = 2HNO_2 (aq) epsilon^o = 0.92 v Derive the reduction potential for reaction 3 above from two reduction potentials [2 above and 4 below], HNO_3 (aq) + 2H^+ (aq) + 2e = HNO_2 (aq) + H_2 O(l) epsilon^o = 0.94...
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 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)...
4. (a) Use the standard reduction potentials at 25° C in Table 18.1 in Tro, Fridgen and Shaw, and calculate the standard emf E° of an electrochemical cell described by the following reaction: 3 Zn + 2 Cr3+ + 2 Cr + 3 Zn? (b) What is n? (c) What is AGº for this reaction at 25°? (d) What is the equilibrium constant for this reaction at 25°? TABLE 18.1 Standard Reduction Potentials at 25°C EV) 2.87 1.61 1.51 1.36...
Based on the information in the table of standard reduction potentials below, what is the standard cell potential for a galvanic cell that has sodium and copper electrodes immersed in 1M Nat and Cu2+ solutions? Also, identify the cathode. Half-reaction E° (V) Aut te - Au +1.69 I2 + 2e - + 21 +0.54 Cu²+ + 2e - Cu +0.34 → Fe -0.04 Feit +3e Zn2+ + 2e -0.76 - Zn -2.71 Natte Na 0 -3.05 V Na is the...
Question 7 Using the table of standard reduction potentials shown below, calculate the standard cell potential for a battery based on the following reactions. • Pet2 + 2e + Fe . 2 Li + 2 Li + 2e Reduction Half-Reaction F2 +2e + 2F MnO + 8H+ Se + Mn+2+ 4 H 0 Cl2 + 2e + 20 02 + 4H' + 4e + 2 H2O Ag+ e -- Ag Fet) + e + Fe2 O2 + 2 H2O +...
Question 5 Using the table of standard reduction potentials shown below, calculate the standard cell potential for a battery based on the following reactions. • O2 + 4H+ + 4e + 2 H20 • 2 Cu + 2 Cu+2 + 4e Reduction Half-Reaction F2 + 2e + 2F MnO, +8 H+ + 5e + Mn+2 +4 H20 Cl2 + 2e → 2C O2 + 4H+ + 4e + 2 H2O Agt! + e + Ag Fet3 + e - Fet2...
Consider the following species. Cut Ce3+ Ag+ Zn2+ What is the standard potential for the reaction of Cut with Zn2+ to produce Cu2+ and Zn? E = 0.28 X v Will Cut be able to reduce Zn2+ to Zn? no (yes or no) What is the standard potential for the reaction of Ce3+ with Ag! to produce Ag? Ex= 0.90 x v Will Cell be able to reduce Ag! to Ag? yes (yer or no) Ered (V) 0.68 0.52 0.40...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery that... Reduction Half-Reaction F2(g) + 2e →2F(aq) S2082 (aq) + 2e- → 25042 (aq) O2(g) + 4H+ (aq) + 4e → 2H2O(1) Br2(1) + 2e + 2Br (aq) Agt(aq) + e → Ag(s) Fe3+ (aq) + e- → Fe2+ (aq) 126) + 2e → 21 (aq) Cu2+ (aq) + 2e → Cu(s) Sn4+ (aq) + 2e → Sn2+ (aq) S(s) + 2H+ (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...