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 | |
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) | − 0.40 | Li+(aq)+e−→Li(s) | − 3.04 |
Use the tabulated electrode potentials to calculate K for the oxidation of zinc by H+:
Zn(s)+2H+(aq)→Zn2+(aq)+H2(g)
D)
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) | − 0.40 | Li+(aq)+e−→Li(s) | − 3.04 |
Use tabulated standard electrode potentials to calculate the standard cell potential for the reaction occurring in an electrochemical cell at 25 ∘C. (The equation is balanced.)
Sn2+(aq)+Cd(s)→Sn(s)+Cd2+(aq)
E)
Balance the following redox reactions occurring in acidic aqueous solution.
MnO−4(aq)+Al(s) → Mn2+(aq)+Al3+(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...
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)...
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...
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)...
Write the half reactions and overall reaction for each cell with calculated overall potentials as shown in Table 5-1. (Note: for the iron solutions the Nernst equation must be used) Pb(s) | Pb(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) Cu(s) Zn(s) | Zn(NO3)2(0.1M) || Cu(NO3)2 (0.1M) Cu(s) Cds) | Ca(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) | Cu(s) Cu() Cu(NO3)2(0.1M) Il Fe (0.1M/Fe? (0.1M graphite Pb(s) Pb(NO3)2(0.1M) Il Fe3(aq) (0.1M)/ Fe2(aq) (0.1MI graphite(s) Zns | Zn(NO3)2 (0.1M) || Pb(NO3)2 (0.1M) | Pb(s) Cdis Ca(NO3)2...
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...
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...
For all of the following experiments, under standard conditions, which species could be spontaneously produced? A lead wire is placed in a solution containing Cu2+ yes no Cu yes no PbO2 yes no No reaction Crystals of I2 are added to a solution of NaCl. yes no I- yes no No reaction yes no Cl2 A silver wire is placed in a solution containing Cu2+ no yes Cu no yes No reaction no yes Ag+ Half-Reaction 8° (V) Half-Reaction 8° (V) 2.87 1.99 1.82 1.78 1.70 1.69 1.68 1.60...
Using the information in the table: Which combination of metals, if used to create an electrochemical cell, would produce the largest voltage? Liu lur the reaction between Zn and Cu2+ ions is 1.1030 V, we can use the known value for the half-cell potential for zinc to determine the half-cell potential for copper: Zn(s) → Zn2+(aq) + 2e + Cu2+(aq) + 2e → Cu(s) Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) E half-cell = 0.7628 V Eºhalf-cell =...
Fill in the Blanks 0.34 Half Reaction E. (V) Half Reaction Erd® (V) F + 2e →2F 0+ 2H + 2e →H,02 0.68 Ag* + e → Ag Cu* + e → Cu 0.52 Co? + e Co? O2 + 2H,0 + 4e → 40H 0.40 H,O, + 2H + 2e → 2H,0 Cu2+ + 2e → Cu Cet+e → Ces Cu2+ + e → Cut 0.16 PbO, + 4H+ +50,2 +2e → PbSO, + 2H,0 2H* + 2e →...
2.87 Ered® (V) 0.68 0.52 0.40 0.34 0.16 Half Reaction F,+ 2e →2F Ag* + e → Ag Co3 + e + CO2- H2O2 + 2H+ + 2e → 2H,0 Ce4+ + e → Ce+ PbO, + 4H+ + SO42- + 2e → PbSO, + 2H,0 Mno, + 4H+ + 3e → MnO2 + 2H,0 2e + 2H+ + 10, → 103 + H2O Mn0, +8H+ + 5e → Mn2+ + 4H,0 Aul+ + 3e → Au Cl2 + 2e...