Consider die electrolysis of molten CuBr_2. The product produced at the cathode

Question

Consider die electrolysis of molten CuBr_2. The product produced at the cathode is d. Br_2 Br Cu^2+ Cu Consider the electrolysis of molten CuBr_2. The product produced at the anode is ____. See question #1 for choices a-d. Consider the electrolysis of molten CsF. The product produced at the anode is ____. F F_2 Cs^+ Cs Consider the electrolysis of molten CsF. The product produced at the cathode is ___. See question #3 for choices a-d. In each case, select the correct half-reaction. Your choices are a-h. Choices a-d are the same for every question. Choices e-h are unique to salt in question. H_2 (g) + 2 OH^- (aq) rightarrow 2 H_2 O + 2 e^- O_2 (g) + 4 H^+ (aq) + 4 e^- rightarrow 2 H_2 O (l) 2 H_2 O + 2 e%^- rightarrow H_2 (g) + 2 OH^- (aq). 2 H_2 O (l) rightarrow O_2 (g) + 4 H^+ (aq) + 4 e^- Consider the electrolysis of an aqueous solution Cu (NO_3)_2. The half-reaction occurring at the anode is d. Cu(s) rightarrow Cu^2+ (aq) + 2 e^- NO_3 (aq) + 4 H^+ (aq) + 3 e^- rightarrow NO(g) + 2 H_2 O (l) Cu^2+ (aq) + 2 e^- rightarrow Cu(s) NO(g) + 2 H_2 O (l) rightarrow NO_3^- (aq) + 4 H^+ (aq) + 3 e^- Consider the electrolysis of an aqueous solution of Cu (NO_3)_2. The half-reaction occurring at the cathode is g. See question #5 for choices e-h. Consider the electrolysis of an aqoeous solution of Kl. The half-reaction occurring at the cathode is ____. K(s) rightarrow K^+ (aq) + e^- K^+ (aq) + e^- rightarrow K(s) 2 I^- (aq) rightarrow I_2 (g) + 2 e^- I_2 (g) + 2 e^- rightarrow 2 I^- (aq) Consider the electrolysis of an aqueous solution of KI. The half-reaction occurring at the anode is ___. See question #7 for choices e-h.

Explanation / Answer

1.CuBr2 (aq)
cathode Cu2+ + 2e- ---> Cu (Reduction)

2.anode 2Br- - 2e- ---> Br2(oxidation)

3.CsF(molten)

Anode:2F^- - 2e-. ------>F2

4.Cathode: Cs------>Cs +e-

5.Electrolysis of Cu(NO3)2(aq) would yield Cu(s) at the cathode ,NO3- cannot be oxidized.

2Cu+2e- –----> 2Cu(s)

6.Electrolysis of a copper(II) nitrate solution produces oxygen at the anode and copper at the cathode. The half reaction is:
2H2O = 4H+(aq) + O2(g)

7.During electrolysis, positive ions are attracted to the cathode. Hence, K+ and H+ ions move to the cathode. Following the ease of discharge list, the less reactive metal always gets discharged(except in some cases where the more reactive is discharged as it is CONCENTRATED) Hence, as Hydrogen is less reactive than Potassium, it gains one electron from the cathode and is discharged as hydrogen gas. Potassium ions still remain in the solution. The equation for reaction at the cathode is (H+ + e- -------> H2) Reactions at cathode is always Reduction(gaining of electrons) Gaining of e- is always written on the left-hand side of the equation.

8.At the anode, I- and OH- ions are attracted to the anode. If it was Concentrated Potassium Iodide instead of Aqueous Potassium Iodide, Iodide ions will get discharged. However, in this case where it is Aqueous, OH- ions are less reactive and they get discharged. They lose four electrons to form water and oxygen.(4OH- ---------->2H2O + O2 + 4e-)
Reactions at anode is always Oxidation(losing electrons) Losing of electrons is always written on the right hand side of the equation.

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