CHEMISTRY 115 HOMEWORK 7 NAME _______________________

CHEMISTRY 115	QUIZ 4
MAY 10, 2001		NAME _____________________________

Multiple Choice: Place the letter corresponding to the best answer in front of each of the following.

(4)

_D__1. The best oxidizing agent in the following list:

	A. Cr	B. Cr³⁺	C. Sn	D. Sn²⁺
	-------	-0.74	---	-0.14 V

_A__2. The best reducing agent in the following list

	A. Cr	B. Cr³⁺	C. Sn	D. Sn²⁺
	+0.74	-----	+0.14	------

_C__3. For the reaction Co(s) + Ni²⁺ à Co²⁺ + Ni(s), E^o_cell= +0.03 V. If cobalt metal is added to a

water solution in which [Ni²⁺] = 1.0 M

The reaction will not proceed in the forward direction at all
The displacement of Ni²⁺ from solution by Co will go to completion
The displacement of Ni²⁺ will proceed to a considerable extent but the reaction will stop before the Ni²⁺ is completely displaced
Only the reverse reaction will occur

_B__4. The reaction Cu²⁺(aq) + 2 Cl^-(aq) à Cu(s) + Cl₂(g) has E^o_cell = -1.02 V. This reaction

can be made to produce electricity in a voltaic cell
can be made to occur in an electrolysis cell
occurs whenever Cu²⁺ and Cl^- are brought together in aqueous solution
can occur in acidic solution but not in basic solution

Write cell notation for the following galvanic cell, If reactants are in solution or are gases assume a platinum electrode. (2)

Br₂(aq) + 2 Fe²⁺(aq) à 2 Br^-(aq) + 2 Fe³⁺

Br₂ is gaining electrons = cathode; Fe²⁺ is losing electrons = anode

Cell notation requires that the anode be written before the cathode.

Need Pt electrodes because there are no metal just aqueous species present.

Pt|Fe²⁺, Fe³⁺ || Br₂(aq), Br^- | Pt

For the reaction 2Ag(s) + Pt²⁺ à 2Ag⁺ + Pt(s), E^o = 0.39 V. What is E^o for the reduction of Pt²⁺ to Pt? (4)

Ag⁺ + e^-= Ag
2(Ag = Ag⁺ + e^-)	e^o = -(0.80 V)
Pt²⁺ + 2 e^- = Pt	e^o = ?
-------------------	----------------------
2Ag(s) + Pt²⁺ à 2Ag⁺ + Pt(s)	e^o = +0.39 V

For Hess' Law type argument, 0.39 V = -0.80 V + e^o_Pt

And e^o_Pt = +0.39 -(-0.80) = 1.19V

OR Using reduction potentials throughout, e^o = e^o_reduction- e^o_Oxidation

0.39 V = e^o_Pt - e^o_Ag = e^o_Pt - 0.80V

E^o = |_1.19 V___

6. Label the sketch of a voltaic cell below showing the anode, the cathode, the signs of the electrodes, and the direction of electron flow. Also give the overall chemical reaction and calculate the standard electrode potential for the cell. (The cell is not necessarily drawn as anode, cathode.)

Ni(s)|Ni²⁺(aq)||Fe²⁺(aq)|Fe(s) or its reverse, i.e. the one which corresponds to the spontaneous reaction.) For a voltaic cell, the emf must be positive. (6)

+ ß e^-

ß e^- -

____________

___

( V)

_______________

___

/ ----

---

-----

---

--\

| |

Ni(s), Ni²⁺(1M)

Fe²⁺(1M), Fe(s)

____________

___

______________

cathode

anode

Ni²⁺ + 2 e^- = Ni	e^o = -0.25 V (reduction, cathode)
~~Fe²⁺ + 2 e^- = Fe~~	~~e^o = -0.44 V~~ but need reverse reaction for overall + emf
Fe = Fe²⁺ + 2 e^-	e^o = +0.44 V (oxidation, anode)
-------------------	-----------------
Ni²⁺ + Fe = Ni + Fe²⁺	e^o = +0.19 V

Note that you could also use e^o = e^o_reduction- e^o_Oxidation = -0.25 V - (-0.44V) = +0.19 V

Eo = |_+0.19 V____

Reaction:__ Ni²⁺ + Fe = Ni + Fe²⁺_______________

7. Complete and balance the following redox reaction using either the ion-electron or oxidation number method. (No credit for by inspection) and then write in the appropriate chemical species requested. (5)

C₂H₅OH(aq) + MnO₄^- --basic-> MnO₂(s) + C₂H₃O₂^-

3[H₂O + C₂H₅OH à C₂H₃O₂^- + 5 H⁺ + 4 e^-]

4[3 e^- + 4 H⁺ + MnO₄^- à MnO₂+ 2 H₂O]

---------------------------------------------------------------------

H⁺ + 3 C₂H₅OH + 4 MnO₄^- à 3 C₂H₃O₂^- + 4 MnO₂ + 5H₂O

but cannot have H⁺ in base

OH^- + H⁺ + 3 C₂H₅OH + 4 MnO₄^- à 3 C₂H₃O₂^- + 4 MnO₂ + OH^- + 5 H₂O

3 C₂H₅OH + 4 MnO₄^- à 3 C₂H₃O₂^- + 4 MnO₂ + OH^- + 4 H₂O

Chemical species oxidized:_C₂H₅OH______, Oxidizing agent:_MnO₄^-_______