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Name: Anode Zno 20. Consider the following standard reduction potentials: 2H*(aq) + 2 Hz(8); E-0.00 V...
Given the following standard reduction potentials choose the cell which will work as a voltaic cell....
Given the following standard reduction potentials choose the cell which will work as a voltaic cell. All cells below are written according to the usual cell diagram convention. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V 2H+(aq) + 2e → H2(g) E° = 0.00 V Sn2+ (aq) + 2e → Sn(s) E° = -0.14 V Ni2+(aq) + 2e → Ni(s) E° = -0.26 V Cd2+(aq) + 2e → → Cd(s) E° = -0.40 V Sn(s) | Sn2+(aq) || Ni2+(aq)...
Exercise 18.15 The voltaic cell is represented with the line notation. Standard Reduction Half-Cell Potentials at...
Exercise 18.15 The voltaic cell is represented with the line notation. Standard Reduction Half-Cell Potentials at 25 °C Half Reaction HNO2 (aq) + 2 H+ (aq) + e--+ NO(g) + H20(1) NO. (aq) + 4 H + (aq) + e-? NO(g) + 2 H2O(1) sn't (aq) + 2 e-? Sn2 + (aq) 2H+ (aq) + 2e-?H2(g) Sn2+ (aq) + 2 e-? Sn(s) E (V) 0.98 0.96 0.15 0.00 -0.14
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and...
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e– → Sn(s). What is the concentration of Sn2+ if Zn2+ is 2.5 × 10−3 M and the cell emf is 0.660 V? The standard reduction potentials are given below Zn+2(aq) + 2 e− → Zn(s) E∘red == −0.76 V Sn2+(aq) + 2 e– → Sn(s) E∘red −0.136 V
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)...
Standard (reduction) potentials are given below for several half-reactions. Use them to answer the following. Cd2+...
Standard (reduction) potentials are given below for several half-reactions. Use them to answer the following. Cd2+ (aq) + 2 e- → Cd(s) Mn2+(aq) + 2 e- → Mn(s) Cu2+ (aq) + 2 e- → Cu(s) Cr3+(aq) + 3 e- → Cr(s) Sn2+(aq) + 2 e- → Sn(s) A13+(aq) + 3 e- → Al(s) Co2+ (aq) + 2 e- → Co(s) Ered = -0.403 V Ered = -1.185 V Ered = +0.342 V Ered = -0.744 V Ered = -0.138 V...
Two standard reduction potentials are given below. Cd2+(aq) + 2 e− → Cd(s) E⁰red = −0.403...
Two standard reduction potentials are given below. Cd2+(aq) + 2 e− → Cd(s) E⁰red = −0.403 V Al3+(aq) + 3 e− → Al(s) E⁰red = −1.662 V There is only ONE submission for each part. (a) Which is a stronger reducing agent, Cd(s) or Al(s)? Cd(s) Al(s) (b) Which is the most difficult to oxidize, Cd(s) or Al(s)? Cd(s) Al(s) (c) Is Cd(s) able to reduce Al3+(aq) in a spontaneous reaction? is able is not able (d) Is Al(s) able...
Standard (reduction) potentials are given below for several half-reactions. Use them to answer the following. Mn2+...
Standard (reduction) potentials are given below for several half-reactions. Use them to answer the following. Mn2+ (aq) + 2e- → Mn(s) Cr3+(aq) + 3 e- → Cr(s) Sn2+(aq) + 2 e--> Sn(s) Cu2+ (aq) + 2 e → Cu(s) A13+ (aq) + 3 e- → Al(s) Co2+(aq) + 2 e- → Co(s) Cd2+ (aq) + 2 e- → Cd(s) Ered = -1.185 V Ered = -0.744 V Ered = -0.138 V = +0.342 V Ered = -1.662 V Ered =...
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...
Answer Cr, Sn, Cu, Co, Cd is wrong. Standard (reduction) potentials are given below for several...
Answer Cr, Sn, Cu, Co, Cd is wrong.
Standard (reduction) potentials are given below for several half-reactions. Use them to answer the following. Mn2+(aq) + 2 е > Mn(s) Ered = -1.185 V Ered Cr3+(ag) + 3е > Cr(s) = -0.744 V Ered Sn2 (aq)2 e = -0.138 V Sn(s) Ered Cu2+(aq) 2 e » Cu(s) = +0.342 V Ered AI3+(aq) + 3 е > = -1.662 V Al(s) Ered =-0.280 v Со2+(aq) + 2 е > Co(s) Cd2+(aq)2 e...
Selective Reduction The standard reduction potential for the half-reaction: Sn4+ + 2e - Sn2+ is +0.15...
Selective Reduction The standard reduction potential for the half-reaction: Sn4+ + 2e - Sn2+ is +0.15 V. Consider data from the table of standard reduction potentials for common half-reactions, in your text. For a galvanic cell under standard conditions, which of the following anodic half reactions would produce, at the cathode a spontaneous reduction of Sn4+ to Sn2+ but not Sn2+ to Sn. no yes yes yes yes yes Fe — Fe2+ + 2e- Sn2+ Sn4+ + 2e- Sn Sn2+...
Given the following standard reduction potentials choose the cell which will work as a voltaic cell. All cells below are written according to the usual cell diagram convention. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V 2H+(aq) + 2e → H2(g) E° = 0.00 V Sn2+ (aq) + 2e → Sn(s) E° = -0.14 V Ni2+(aq) + 2e → Ni(s) E° = -0.26 V Cd2+(aq) + 2e → → Cd(s) E° = -0.40 V Sn(s) | Sn2+(aq) || Ni2+(aq)...
Exercise 18.15 The voltaic cell is represented with the line notation. Standard Reduction Half-Cell Potentials at 25 °C Half Reaction HNO2 (aq) + 2 H+ (aq) + e--+ NO(g) + H20(1) NO. (aq) + 4 H + (aq) + e-? NO(g) + 2 H2O(1) sn't (aq) + 2 e-? Sn2 + (aq) 2H+ (aq) + 2e-?H2(g) Sn2+ (aq) + 2 e-? Sn(s) E (V) 0.98 0.96 0.15 0.00 -0.14
Standard (reduction) potentials are given below for several half-reactions. Use them to answer the following. Cd2+ (aq) + 2 e- → Cd(s) Mn2+(aq) + 2 e- → Mn(s) Cu2+ (aq) + 2 e- → Cu(s) Cr3+(aq) + 3 e- → Cr(s) Sn2+(aq) + 2 e- → Sn(s) A13+(aq) + 3 e- → Al(s) Co2+ (aq) + 2 e- → Co(s) Ered = -0.403 V Ered = -1.185 V Ered = +0.342 V Ered = -0.744 V Ered = -0.138 V...
Standard (reduction) potentials are given below for several half-reactions. Use them to answer the following. Mn2+ (aq) + 2e- → Mn(s) Cr3+(aq) + 3 e- → Cr(s) Sn2+(aq) + 2 e--> Sn(s) Cu2+ (aq) + 2 e → Cu(s) A13+ (aq) + 3 e- → Al(s) Co2+(aq) + 2 e- → Co(s) Cd2+ (aq) + 2 e- → Cd(s) Ered = -1.185 V Ered = -0.744 V Ered = -0.138 V = +0.342 V Ered = -1.662 V Ered =...
Answer Cr, Sn, Cu, Co, Cd is wrong.
Standard (reduction) potentials are given below for several half-reactions. Use them to answer the following. Mn2+(aq) + 2 е > Mn(s) Ered = -1.185 V Ered Cr3+(ag) + 3е > Cr(s) = -0.744 V Ered Sn2 (aq)2 e = -0.138 V Sn(s) Ered Cu2+(aq) 2 e » Cu(s) = +0.342 V Ered AI3+(aq) + 3 е > = -1.662 V Al(s) Ered =-0.280 v Со2+(aq) + 2 е > Co(s) Cd2+(aq)2 e...
Selective Reduction The standard reduction potential for the half-reaction: Sn4+ + 2e - Sn2+ is +0.15 V. Consider data from the table of standard reduction potentials for common half-reactions, in your text. For a galvanic cell under standard conditions, which of the following anodic half reactions would produce, at the cathode a spontaneous reduction of Sn4+ to Sn2+ but not Sn2+ to Sn. no yes yes yes yes yes Fe — Fe2+ + 2e- Sn2+ Sn4+ + 2e- Sn Sn2+...
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