1. Balance the following reactions and determine the reducing agent and oxidizin

Question

1. Balance the following reactions and determine the reducing agent and oxidizing agent for each reaction.

____N2H4 + ____ClO3- + ____-OH à ____NO3- + ____Cl- + ____H2O

____NaClO3 + ____H2O + ____I2 ? ____H+ + ____IO3- + ____NaCl

2. Which reaction and statement correctly describes the electro-chemical cell shown and what is the voltage that is produced?

a. Cu2+ (aq)     +          2Ag+ (aq)        à Cu (s)         +          2Ag (s)

b. Cu(s)         +          2Ag (s)            à Cu2+ (aq)       +          2Ag+ (aq)       

c. Cu2+ (aq)   +          2Ag (s)            à Cu (s)         +          2Ag+ (aq)       

d. Cu(s)         +          2Ag+ (aq)        à Cu2+ (aq)    +          2Ag (s)

a. e’s flow from the Cu cathode to the Ag anode and K+ ions diffuse into the AgNO3 solution.

b. e’s flow from the Cu cathode to the Ag anode and K+ ions diffuse into the Cu(NO3)2 solution.

c. e’s flow from the Cu anode to the Ag cathode and K+ ions diffuse into the Cu(NO3)2 solution.

d. e’s flow from the Cu anode to the Ag cathode and K+ ions diffuse into the AgNO3 solution.

a. +1.14 V

b. +0.46 V

c. +1.48 V

d. -1.14 V

e. -0.46 V       

3. Calculate Ecell for the following voltaic cell.

Ag / Ag+(1.0 × 10-5M) // Au3+ (1.0 × 10-1M) / Au

a. +0.78 V                   b. +0.46 V                   c. +0.98 V                   d. +2.58 V                   e. -0.78 V       

4. A reactant–favored electrochemical reaction has:

a. ?G° = 0, E° = 0, and K >> 1

b. ?G° < 0, E° > 0, and K > 1

c. ?G° > 0, E° < 0, and K < 1

d. ?G° > 0, E° < 0, and K > 1

e. ?G° < 0, E° = 0, and K >> 1

Explanation / Answer

1. Balanced equations

(a) 3N2H4 + 7ClO3- + 6OH- ---> 6NO3- + 7Cl- + 9H2O

Oxidizing agent = ClO3-

Reducing agent = N2H4

(b) 5NaClO3 + 3H2O + 3I2 ---> 6H+ + 6IO3- + 5NaCl

oxidizing agent = NaClO3

reducing agent = I2

2. The correct cell reaction for the given electrochemical cell would be,

Cu(s) + 2Ag+(aq) ---> Cu2+(aq) + 2Ag(s)

Cell potential = 0.34 - 0.80 = -0.46 V

e. -0.46 V

Correct statement would be,

c. e’s flow from the Cu anode to the Ag cathode and K+ ions diffuse into the Cu(NO3)2 solution.

3. Ecell for the given cell reaction,

Ecell = (1.5 - 0.8) - 0.0592/3 log(1 x 10^-5)^3/(1 x 10^-1)

         = 0.98 V

c. 0.98 V

4. A reactant favoured electrochemical cell would be,

c. dGo > 0, Eo < 0, and K < 1

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