Soal-Soal Latihan Redoks dan Jawabannya

Oxidation – Reduction Reactions

Level 2

Serial Response

For the next 4 questions:

A solution of lead bromide is placed in an electrolytic cell. An electrical current is

passed through the solution.

414. Name all the substances with significant concentrations in this solution.

ANS: Pb⁺²,  Br^- &  HOH.

415. Which substance is most readily formed at the anode?  Explain why.

ANS: Br^-; more likely to oxidize than HOH.

416.Write an electronic equation for the half-reaction taking place at the cathode.

ANS: Pb⁺²(aq)  + 2 e^-  à  Pb(s)

417.  Write the complete redox equation and find how much voltage must be supplied by

    a generator to make this reaction happen.   Show reasoning for this answer.

ANS: Pb⁺²(aq) + 2 e^-  à  Pb(s);   ⁰E  =  -0.13 volt

          2Br^-(aq)  à  Br₂(g)  +  2e^-; ⁰E  = -1.06 volts

          Pb⁺² + 2Br^-  à  Pb + Br₂↑;  ⁰E  = -1.19 volts

         Therefore supplied voltage must be > +1.19 volts.

For the next 5 questions:

A solution of copper(II) sulfate is placed in an electrolytic cell.  An electrical current is

passed through the solution.

418. Name all the substances in this solution that have significant concentrations.

ANS: Cu⁺²,  SO₄^-2 &  HOH.

419. Write an electronic equation for the half-reaction taking place at the cathode.

ANS: Cu⁺²(aq)  + 2 e^-  à  Cu(s)

420. Which substance is most readily formed at the anode?  Explain why.

ANS: O₂; HOH is only substance capable of losing electrons.

421. Balance the full redox equation and determine how much voltage must be

    supplied by a generator to make this reaction go.

ANS: Cu⁺²(aq)  + 2 e^-  à  Cu(s);                        ⁰E  =  +0.34 volt

          HOH(l)  à  ½ O₂(g)  +  2 H⁺(aq)  +  2 e^-; ⁰E  =  -1.23 volts

         Cu⁺² + HOH  à  Cu  +  ½ O₂↑  +  2H⁺;     ⁰E  =  -0.89 volt

     

    Therefore supplied voltage must be > +0.89 volt.

422. Name the valuable substances produced in this electrolytic cell.

ANS: Cu(s),  O₂(g)  &  H₂SO₄

The next 4 questions involve a solution of aluminum fluoride which is undergoing electrolysis.

423. Write the equation for the ionization of aluminum fluoride.

ANS: AlF₃(s)  à  Al⁺³(aq)  +  3 F^-(aq)

424. What compound remains in the solution when the electrolytic cell is in operation?  Why?

ANS: AlF₃; Al⁺³ & F^- are least likely to reduce, oxidize respectively.

425. Write the complete reaction taking place in this electrolytic cell.

ANS: 2 HOH(l)  + 2 e^-  à  2 OH^-(aq)  +  H₂(g)

           HOH(l)  à  ½ O₂(g)  +  2 H⁺(aq)  + 2 e^-

          3HOH(l)  à  2HOH(l) + H₂(g)↑ + ½ O₂(g)↑

426. How much voltage must be supplied by a generator to make this reaction spontaneous?

ANS: E⁰  =  (-0.83 volt)  +  (-1.23 volts)  =  -2.06 volts

          Therefore, supplied voltage must be > +2.06 volts.

For the next 5 questions:

A chemical cell for the decomposition by electrolysis is set up.  Aluminum bromide

is dissolved in water to form an aqueous solution of aluminum bromide.  A table of

reduction potentials is needed to answer the following.

427. List the substances with significant concentrations present in this solution.

ANS: Al⁺³,  Br^-  &  HOH.

428. Which substance is most readily formed at the cathode?  Explain why.

ANS: 2 HOH(l)  +  2 e^-  à  2 OH^-(aq)  +  H₂(g)↑; H₂ has gained electrons.

429. What remains in the solution during electrolysis?

ANS: Al(OH)₃

430. Write an electronic equation for the half-reaction taking place at the anode.

ANS: 2 Br^-(aq)  à  Br₂(g)↑  +  2 e^-

431. Write a balanced equation giving the complete results of the electrolysis of this solution.

ANS: 2 AlBr₃  +  6 H₂O  à  2 Al(OH)₃  +  3 H₂(g)↑  +  3 Br₂(g)↑

The next 4 questions  deal with a solution of nickel iodide; it is placed in an

electrolytic cell and current is supplied.

432. List all the substances with significant concentrations in this solution.

ANS: Ni⁺²,  I^-  &  HOH.

433. Which substance is most readily formed at the cathode?  Explain why.

ANS: Ni⁺²; more reducible than HOH & I^-.

434. What substance remains behind during electrolysis? Why?

ANS: HOH; Ni⁺² & I^- are attracted to the cathode and anode respectively.

435. Balance the full redox reaction and calculate the voltage needed to make this

    electrolytic cell to be effective.

ANS: Ni⁺²(aq)  +  2 e^-  à  Ni(s); ⁰E  =  -0.25 volt

          2 I^-(aq)  à  I₂(s)  +  2 e^-;    ⁰E  =  -0.53 volt

          Ni⁺²  +  2I^-  à  Ni  + I₂;      ⁰E  =  -0.78 volt

          Therefore, a generator must supply voltage > +0.78 volt.