Classify the half‑reactions as reduction half‑reactions or oxidation half‑reactions.
H2(g)⟶2H+(aq)+2e−H2(g)⟶2H+(aq)+2e−
12O2(g)+2H+(aq)+2e−⟶H2O(g)12O2(g)+2H+(aq)+2e−⟶H2O(g)
Cd(s)+2OH−(aq)⟶Cd(OH)2(s)+2e−Cd(s)+2OH−(aq)⟶Cd(OH)2(s)+2e−
2NiO(OH)(s)+2H2O(l)+2e−⟶2Ni(OH)2(s)+2OH−(aq)2NiO(OH)(s)+2H2O(l)+2e−⟶2Ni(OH)2(s)+2OH−(aq)
Fe(s)⟶Fe2+(aq)+2e−Fe(s)⟶Fe2+(aq)+2e−
oxidation
reduction
reduction
oxidation
reduction
H2(g) ---> 2H+ + 2e-
This is an oxidation half reaction as here electrons are being released in the reaction.
O2 + 2H+ + 2e- --> H2O
This is a reduction half reaction as here electrons are being consumed in the product formation.
Cd(s) + 2OH-(aq) --> Cd(OH)2 +2e-
This is an oxidation half reaction as here electrons are released in the product formation.
2NiO(OH)(s)+2H2O(l)+2e−⟶2Ni(OH)2(s)+2OH−(aq)
This is a reduction half reaction as here electrons are being consumed in the product formation.
Fe(s)⟶Fe2+(aq)+2e−
This is an oxidation half reaction as here electrons are being released in the product formation.
Classify the half‑reactions as reduction half‑reactions or oxidation half‑reactions. H2(g)⟶2H+(aq)+2e−H2(g)⟶2H+(aq)+2e− 12O2(g)+2H+(aq)+2e−⟶H2O(g)12O2(g)+2H+(aq)+2e−⟶H2O(g) Cd(s)+2OH−(aq)⟶Cd(OH)2(s)+2e−Cd(s)+2OH−(aq)⟶Cd(OH)2(s)+2e− 2NiO(OH)(s)+2H2O(l)+2e
Classify the half-reactions as reduction half-reactions or oxidation half-reactions. H(g)2 H (aq) +2e Answer Bank 0, (8)+2H (aq)+2e H,O(g) oxidation reduction Cd(s)+20H (aq) Cd(OH), (s) + 2 e 2 NiO (OH)(s)+2H,O(1)+2 e 2 Ni(OH), (s)+ 2 OH (aq) Fe(s)Fe (aq) +2 e
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...
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.
Standard Electrode Potentials at 25?C Reduction Half-Reaction E?(V) F2(g)+2e? ?2F?(aq) 2.87 Au3+(aq)+3e? ?Au(s) 1.50 Cl2(g)+2e? ?2Cl?(aq) 1.36 O2(g)+4H+(aq)+4e? ?2H2O(l) 1.23 Br2(l)+2e? ?2Br?(aq) 1.09 NO3?(aq)+4H+(aq)+3e? ?NO(g)+2H2O(l) 0.96 Ag+(aq)+e? ?Ag(s) 0.80 I2(s)+2e? ?2I?(aq) 0.54 Cu2+(aq)+2e? ?Cu(s) 0.16 2H+(aq)+2e? ?H2(g) 0 Cr3+(aq)+3e? ?Cr(s) -0.73 2H2O(l)+2e? ?H2(g)+2OH?(aq) -0.83 Mn2+(aq)+2e? ?Mn(s) -1.18 How can the table be used to predict whether or not a metal will dissolve in HCl? In HNO3? Drag the terms on the left to the appropriate blanks on the right to...
Part III: Oxidation and Reduction of H2O2 1. Reduction of H2O2: H2O2(aq) + Cr(OH)3(S) (BASIC) • Observations · Evidence of the oxidation of Cr3+ • Reduction Half Reaction • Oxidation Half Reaction • Overall Balanced Redox Reaction · Explain occurrence or non-occurrence of reaction by calculating cell. 2. Oxidation of H2O2: H2O2(aq) + FeCl3(aq) • Observations • Evidence of the oxidation of H2O2 • Reduction Half Reaction • Oxidation Half Reaction • Overall Balanced Redox Reaction • Explain occurrence or...
Classify the following as acid-base reactions or oxidation-reduction reactions. (a) 3 HClO4(aq) + Fe(OH)3(s) → Fe(ClO4)3(aq) + 3 H2O(l) (b) HC2H3O2(aq) + NaHCO3(aq) → NaC2H3O2(aq) + CO2(g) + H2O(l) (c) 4 H2O(l) + 2 KMnO4(aq) → 2 MnO2(s) + 2 KOH(aq) + 3 H2O2(aq) (d) Cl2(g) + KOH(aq) → KClO(aq) + HCl(aq) (e) 2 Fe(s) + 3 NaOCl(aq) → Fe2O3(s) + 3 NaCl(aq) (f) CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(s) Explain your reasoning
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)...
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 rechargeable nickel-cadmium (NiCd) battery contains the following half-reactions: NiO2 +2H2O(l)+2e– → Ni(OH)2 +2OH– E0 =0.49V Cd(OH)2 +2e– →Cd(s)+2OH– E0 =–0.81V a. What is the standard cell potential or voltage of this NiCd cell? b. Write the net chemical reaction in the direction of spontaneous reaction. Is cadmium oxidized or reduced? c. Write an expression for the reaction quotient Q. What is Q if the electrolyte concentrations are: [NiO2] = 1 M and [Cd(OH)2] = 0.01M and [Ni(OH)2] = 0.001...
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...