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94. The net ionic equation for the reaction, Cu(s) + 2HNO3(aq) - H2(g) + Cu(NO3)(aq) A)...
PQ-19. What is the balanced net ionic equation for the reaction of CuCl(aq) and H2(g)? (A)...
PQ-19. What is the balanced net ionic equation for the reaction of CuCl(aq) and H2(g)? (A) Cu?+(aq) + H2(g) → Cu(s) + 2H*(aq) CuCl2 + Ha > H (B) CuCl2(aq) + H2(g) → Cu(s) + 2HCl(aq) (C) Cu?"(aq) + Cl2(aq) + H2(g) → Cu(s) + 2H*(aq) + 2Cl(aq) cu +261 + 2H+ $ (D) Cu2+(aq) + 2CH(aq) + H2(g) → Cu(s) + 2H+ (aq) + 2CH(aq)
1. For the oxidation-reduction reaction Zn(s) + 2HNO3 (aq) → H2(g) + Zn(NO3)2 (aq) Hint: HNO3...
1. For the oxidation-reduction reaction Zn(s) + 2HNO3 (aq) → H2(g) + Zn(NO3)2 (aq) Hint: HNO3 a) provide equation for the oxidation half-reaction: b) provide equation for the reduction half-reaction: In the given equation above or in your answers to a) and b): c) label the substance which is oxidizing agent, d) label the substance which is reducing agent.
Consider the unbalanced chemical reaction shown below: Cu(s)+HNO3(aq)→Cu(NO3)2(aq)+NO(g)+H2O(l) The oxidation state of Cu in Cu(s)Cu(s) =...
Consider the unbalanced chemical reaction shown below: Cu(s)+HNO3(aq)→Cu(NO3)2(aq)+NO(g)+H2O(l) The oxidation state of Cu in Cu(s)Cu(s) = The oxidation state of Cu in Cu(NO_3)_2(aq)Cu(NO3)2(aq) = The oxidation state of N in HNO_3(aq)HNO3(aq) = The oxidation state of N in NO(g)NO(g) = The total number of electrons transferred in this reaction is = The sum of the coefficients in the balanced chemical reaction =
Question 6 Calculate K by using the electrode potentials for the following equation: Cu(s) + 2H+(aq)...
Question 6 Calculate K by using the electrode potentials for the following equation: Cu(s) + 2H+(aq) —> Cu2+(aq) + H2(g) K- X10
39. Given: E +0.46 V Eo- +0.34 V Cu(s) + 2 Ag+ → 2 Ag (s) + Cu2. Cu2+ + H2 (g) → Cu (s) + 2 H' Find the standar...
39. Given: E +0.46 V Eo- +0.34 V Cu(s) + 2 Ag+ → 2 Ag (s) + Cu2. Cu2+ + H2 (g) → Cu (s) + 2 H' Find the standard potential for the cell reaction for 2 Ag + H2 (g)-2 Ag+2H +0.80 V b. a. +0.40 V +0.12 V d. c. -0.12 v none of these e. 40. Given: Fe (s) + 2 Ag' (aq) Fe (ag)+ 2 Ag (s) with E ell -1.24 V What is the...
*A copper, Cu(s), electrode is immersed in a solution that is 1.00 M in ammonia, NH3, and 1.00 M in tetraamminecopper(II), [Cu(NH3)4]2+. If a standard hydrogen electrode is used as the cathode, the ce...
*A copper, Cu(s), electrode is immersed in a solution that is
1.00 M in ammonia, NH3, and 1.00 M in tetraamminecopper(II),
[Cu(NH3)4]2+. If a standard hydrogen electrode is used as the
cathode, the cell potential, Ecell, is found to be 0.070 V at 298
K.
A copper, Cu(s), electrode is immersed in a solution that is 1.00 M in ammonia, NH3, and 1.00 M in tetraamminecopper(II), [NH. If a standard hydrogen electrode is used as the cathode, the cell potential,...
NO−3(aq)+4H+(aq)+3e−→NO(g)+2H2O(l) E∘=0.96V ClO2(g)+e−→ClO−2(aq) E∘=0.95V Cu2+(aq)+2e−→Cu(s) E∘=0.34V 2H+(aq)+2e−→H2(g) E∘=0.00V Pb2+(aq)+2e−→Pb(s) E∘=−0.13V Fe2+(aq)+2e−→Fe(s) E∘=−0.45V You may want to...
NO−3(aq)+4H+(aq)+3e−→NO(g)+2H2O(l) E∘=0.96V ClO2(g)+e−→ClO−2(aq) E∘=0.95V Cu2+(aq)+2e−→Cu(s) E∘=0.34V 2H+(aq)+2e−→H2(g) E∘=0.00V Pb2+(aq)+2e−→Pb(s) E∘=−0.13V Fe2+(aq)+2e−→Fe(s) E∘=−0.45V You may want to reference (Pages 898 - 902) Section 19.4 while completing this problem. Part A Use data from the table above to calculate E∘cell for the reaction. Fe(s)+2H+(aq)→Fe2+(aq)+H2(g) Express your answer using two decimal places.
Select the pair of reactants that will react. H2(g) + H+(aq) MnCl2(aq) + Cr(s) – AlCl3(aq)...
Select the pair of reactants that will react. H2(g) + H+(aq) MnCl2(aq) + Cr(s) – AlCl3(aq) + Cu(s) – Cu2+ (aq) + Ag(s) - AgCl(aq) + Sn(s) -
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...
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)...
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)...
PQ-19. What is the balanced net ionic equation for the reaction of CuCl(aq) and H2(g)? (A) Cu?+(aq) + H2(g) → Cu(s) + 2H*(aq) CuCl2 + Ha > H (B) CuCl2(aq) + H2(g) → Cu(s) + 2HCl(aq) (C) Cu?"(aq) + Cl2(aq) + H2(g) → Cu(s) + 2H*(aq) + 2Cl(aq) cu +261 + 2H+ $ (D) Cu2+(aq) + 2CH(aq) + H2(g) → Cu(s) + 2H+ (aq) + 2CH(aq)
1. For the oxidation-reduction reaction Zn(s) + 2HNO3 (aq) → H2(g) + Zn(NO3)2 (aq) Hint: HNO3 a) provide equation for the oxidation half-reaction: b) provide equation for the reduction half-reaction: In the given equation above or in your answers to a) and b): c) label the substance which is oxidizing agent, d) label the substance which is reducing agent.
Question 6 Calculate K by using the electrode potentials for the following equation: Cu(s) + 2H+(aq) —> Cu2+(aq) + H2(g) K- X10
39. Given: E +0.46 V Eo- +0.34 V Cu(s) + 2 Ag+ → 2 Ag (s) + Cu2. Cu2+ + H2 (g) → Cu (s) + 2 H' Find the standard potential for the cell reaction for 2 Ag + H2 (g)-2 Ag+2H +0.80 V b. a. +0.40 V +0.12 V d. c. -0.12 v none of these e. 40. Given: Fe (s) + 2 Ag' (aq) Fe (ag)+ 2 Ag (s) with E ell -1.24 V What is the...
*A copper, Cu(s), electrode is immersed in a solution that is
1.00 M in ammonia, NH3, and 1.00 M in tetraamminecopper(II),
[Cu(NH3)4]2+. If a standard hydrogen electrode is used as the
cathode, the cell potential, Ecell, is found to be 0.070 V at 298
K.
A copper, Cu(s), electrode is immersed in a solution that is 1.00 M in ammonia, NH3, and 1.00 M in tetraamminecopper(II), [NH. If a standard hydrogen electrode is used as the cathode, the cell potential,...
Select the pair of reactants that will react. H2(g) + H+(aq) MnCl2(aq) + Cr(s) – AlCl3(aq) + Cu(s) – Cu2+ (aq) + Ag(s) - AgCl(aq) + Sn(s) -
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