Under standard conditions, Consider the following standard reduction potentials, Ni2+(aq) + 2 e- → Ni(s) E°...

Under standard conditions, Consider the following standard reduction potentials,

Ni2+(aq) + 2 e- → Ni(s) E° = -0.26 V I2(s) + 2 e- → 2 I-(aq) E° = +0.54 V Under standard conditions,

a)Ni2+(aq) is a stronger oxidizing agent than I2(s) and I-(aq) is a stronger reducing agent than Ni(s).

b)I2(s) is a stronger oxidizing agent than Ni2+(aq) and Ni(s) is a stronger reducing agent than I-(aq).

c)I-(aq) is a stronger oxidizing agent than Ni(s) and I2(s) is a stronger reducing agent than Ni2+(aq).

d)Ni(s) is a stronger oxidizing agent than I-(aq) and Ni2+(aq) is a stronger reducing agent than I2(s).

Expert Solution

queen_honey_blossom answered 2 years ago

Part A Calculate the equilibrium constant at 25 ∘C for the reaction Ni(s)+2Ag+(aq)→Ni2+(aq)+2Ag(s), if Ni2+(aq)+2e−→Ni(s), E∘...

Part A Calculate the equilibrium constant at 25 ∘C for the reaction Ni(s)+2Ag+(aq)→Ni2+(aq)+2Ag(s), if Ni2+(aq)+2e−→Ni(s), E∘ = -0.26 V, Al+(aq)+e−→Al(s), E∘ = 0.80 V Express your answer using one significant figure. Part B Calculate the equilibrium constant at 25 ∘C for the reaction Hg2+2(aq)→Hg(l)+Hg2+(aq) See Appendix D for standard reduction potentials. Express your answer using one significant figure.

Given the following standard reduction potentials: Pb2+ (aq) +2e- ---> Pb (s) E= -.126V PbSO4(s) +...

Given the following standard reduction potentials: Pb2+ (aq) +2e- ---> Pb (s) E= -.126V PbSO4(s) + 2e- ---> Pb(s) + SO42- (aq) E= -.356V Determine the Ksp for PbSO4(s) at 25 degrees C

For the following reaction: Ni2+(aq) + Co(s) ⇌ Ni(s) + Co2+(aq) a) Calculate E0 (E^0) b)...

For the following reaction: Ni2+(aq) + Co(s) ⇌ Ni(s) + Co2+(aq) a) Calculate E0 (E^0) b) Calculate G0 (delta G^0) c)Calculate K d) If you start with 10.0 g of Ni and 10.0 g of Co in 100.0 mL solution (containing 1.0 M solution of CoCl2 and 0.0000100 M solution of NiCl2 ) which way the reaction will move toward in order to reach equilibrium? e) Calculate the G (delta G) for the reaction in part d. f)Calculate the equilibrium...

Calculate ΔG° for the following reaction as written Ni2+ (aq) + Sn (s) → Ni (s)...

Calculate ΔG° for the following reaction as written Ni2+ (aq) + Sn (s) → Ni (s) + Sn2+ (aq) given the following reduction half-reactions and standard reduction potentials. Sn2+ (aq) + 2 e- → Sn (s) E° = -0.15V Ni2+ (aq) + 2 e- → Ni (s) E° = -0.25V

Calculate the standard cell potential for each of the following electrochemical cells. Part A Ni2+(aq)+Mg(s)→Ni(s)+Mg2+(aq) Express...

Calculate the standard cell potential for each of the following electrochemical cells. Part A Ni2+(aq)+Mg(s)→Ni(s)+Mg2+(aq) Express your answer using two decimal places. E∘cell = _____ V Part B 2H+(aq)+Fe(s)→H2(g)+Fe2+(aq) Express your answer using two decimal places E∘cell =_______ V Part C 2NO−3(aq)+8H+(aq)+3Cu(s)→2NO(g)+4H2O(l)+3Cu2+(aq) Express your answer using two decimal places. E∘cell = ________V

Consider the following half reactions at 298 K Ni2+ + 2 e- → Ni Eo =...

Consider the following half reactions at 298 K Ni2+ + 2 e- → Ni Eo = -0.231 V Pb2+ + 2 e- → Pb Eo = -0.133 V A galvanice cell based on these half reactions is set up under standard conditions where each solution is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Pb electrode weigh when the nonstandard potential of the cell is 0.09296 V?

Calculate the equilibrium constant for the reaction between Ni2+(aq) and Zn(s) under standard conditions at 25∘C.

Half-reaction E° (V) Hg2+(aq) + 2e- -----> Hg(l) 0.855V Ni2+(aq) + 2e- -----> Ni(s) -0.250V Zn2+(aq)...

Half-reaction E° (V) Hg2+(aq) + 2e- -----> Hg(l) 0.855V Ni2+(aq) + 2e- -----> Ni(s) -0.250V Zn2+(aq) + 2e- -----> Zn(s) -0.763V (1) The weakest oxidizing agent is: ___ enter formula (2) The strongest reducing agent is: ___ (3) The strongest oxidizing agent is:___ (4) The weakest reducing agent is: ___ (5) Will Zn(s) reduce Hg2+(aq) to Hg(l)? _____(yes)(no) (6) Which species can be oxidized by Ni2+(aq)? ___ If none, leave box blank.

Reduction half-reaction E∘ (V) Ag+(aq)+e−→Ag(s) 0.80 Cu2+(aq)+2e−→Cu(s) 0.34 Sn4+(aq)+4e−→Sn(s) 0.15 2H+(aq)+2e−→H2(g) 0 Ni2+(aq)+2e−→Ni(s) −0.26 Fe2+(aq)+2e−→Fe(s) −0.45...

Reduction half-reaction E∘ (V) Ag+(aq)+e−→Ag(s) 0.80 Cu2+(aq)+2e−→Cu(s) 0.34 Sn4+(aq)+4e−→Sn(s) 0.15 2H+(aq)+2e−→H2(g) 0 Ni2+(aq)+2e−→Ni(s) −0.26 Fe2+(aq)+2e−→Fe(s) −0.45 Zn2+(aq)+2e−→Zn(s) −0.76 Al3+(aq)+3e−→Al(s) −1.66 Mg2+(aq)+2e−→Mg(s) −2.37 1) Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 ∘C) for the following reaction: Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s) 2) Calculate the standard cell potential (E∘) for the reaction X(s)+Y+(aq)→X+(aq)+Y(s) if K = 3.80×10−4. Express your answer to three significant figures and include the appropriate units.

A voltaic cell is assembled under standard conditions based on the following reaction : Ni(s) +...

A voltaic cell is assembled under standard conditions based on the following reaction : Ni(s) + 2 Ag+(aq) → Ni+2(aq) + 2 Ag(s) Indicate which statements, if any, about this voltaic cell are true by checking the box in front of each true statment. - The voltaic cell will always produce a positive potential. If the [Ni2+] is doubled while [Ag+] remains standard, the intial potential produced by the voltaic cell will double. If the [Ag+] is doubled while [Ni+2]...

Question