CHEMISTRY 115 HOMEWORK 7 NAME _______________________

CHEMISTRY 115 HOMEWORK 5 NAME _______________

Due MAY 9, 01 COMP HWK: Electro: 1-4

For the following reaction:

__ K2Cr2O7 + __ FeCl2 + __ HCl -> __ CrCl3 + __ H2O + __ FeCl3 + __ KCl

Write the net ionic equation K⁺ and Cl^- are the spectator ions

Cr₂O₇^2- + Fe²⁺ --acid-> Cr³⁺ + Fe³⁺

balance_

6 e^- + 6 H⁺ + 8 HCl + K₂Cr₂O₇ à 2 CrCl₃ + 2 KCl + 7 H₂O

6[HCl + FeCl₂ à FeCl₃ + H⁺ + e^-]

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14 HCl + K₂Cr₂O₇ + 6 FeCl₂ à 2 CrCl₃ + 2 KCl + 7 H₂O + 6 FeCl₃

6 e^- + 14 H⁺ + Cr₂O₇^2- à 2 Cr³⁺ + 7 H₂O

6[Fe²⁺ à Fe³⁺ + e^-]

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14 H⁺ + Cr₂O₇^2- + 6 Fe²+ à 2 Cr³⁺ + 7 H₂O + 6 Fe³⁺

Write out as full compounds and balance the K’s and Cl’s

Chemical species oxidized: loses e^- = Fe²⁺ = FeCl₂, Oxidizing agent: gains e⁺ = Cr₂O₇^2- = K₂Cr₂O₇

2. Write the half reactions and a balanced reaction for the following galvanic cell.

Fe (s)| Fe⁺² (aq)|| Br₂(aq), Br^-(aq) | Pt(s)

Fe(s) = Fe²⁺ + 2 e^-

Br₂ + 2 e^- = 2 Br^-

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Fe(s) + Br₂ = Fe²⁺ + 2 Br^-

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

NiO₂(s) + 4H⁺(aq) + 2Ag(s) à Ni⁺²(aq) + H₂O + 2Ag⁺

Ag(s) | Ag⁺ || NiO₂(s), Ni²⁺_.,H⁺ | Pt(s)

4a. Circle the better oxidizing agent in the following pairs at standard conditions.

More easily reduced = higher (more positive) on the reduction emf table.

I. Fe3+ or Pb2+	II. I2 or Sn2+	III. MnO₄^- or Cr₂O₇^2-
+0.77 vs -0.13	+0.54 vs – 0.14	+1.49 (1.507) vs +1.33

b. Circle the better reducing agent in the following pairs at standard conditions.

More easily oxidized = lower (more negative) on the reduction emf table on right side.

I. H2 or I-	II. Sn2+ or Fe2+	III. Sn2+ or Cu
0.0 vs –(+0.54)	-(+0.15) vs –(+0.77)	-(+0.15) vs –(+0.34)

5. 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.)

Pt|Br2(l)|Br-(aq)||Cl-(aq)|Cl2(g)|Pt or its reverse, i.e. the one which corresponds to the spontaneous reaction. For a voltaic or galvanic cell, the overall emf must be positive

e^-à

e^- à

____________

___

( V)

_______________

___

/ ----

---

-----

---

--\

| |

Br₂(l), Br^-(1M)

Cl₂(1atm),Cl^-(1M)

____________

___

_______________

Oxidation =

anode

Reduction =

cathode

Cl₂ + 2 e^- = 2 Cl^-	e⁰ = +1.36 V
2 Br^- = Br₂ + 2 e^-	e^o = -(+1.07 v)
	----------------
	e⁰ = 0.29 V

Eo = |_0.29 V_____

Reaction:__Cl₂ + 2 Br^- = 2 Cl^- + Br₂_________________________________

What is the standard cell potential and the net reaction in a galvanic cell that has the following half reactions? For a voltaic or galvanic cell, the overall emf must be positive

	MnO₂ + 4H⁺ + 2e^{- ó} Mn⁺² + 2H₂O	Eo = 1.23 V
	PbCl₂ + 2e^- ó Pb + 2Cl^-	Eo = -0.27 V

Pb reaction must be reversed for the oxidation and give a + emf

MnO₂ + 4H⁺ + 2e^{- ó} Mn⁺² + 2H₂O	Eo = 1.23 V
Pb + 2 Cl^- = PbCl₂ + 2 e^-	E^o = +0.27 V
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MnO₂ + 4 H⁺ + Pb + 2 Cl^- = Mn²⁺ + 2 H₂O + PbCl₂	E^o = 1.50 V

Eo = |_1.50 V______

The standard voltage for the reaction HClO(aq) + H+(aq) + 2 Cr2+(aq) = 2 Cr3+(aq) + Cl-(aq) + H2O(l) is Eo = 1.80V. Use appropriate data from the EMF table in Appendix E-8 to calculate Eo for the half reaction HClO(aq) + H+(aq) + 2 e- = Cl-(aq) + H2O(l).

HClO(aq) + H+(aq) + 2 e^- = Cl^-(aq) + H2O(l)	e^o = ?
Reverse 2[Cr³⁺ + e^- = Cr²⁺]	e^o = -(-0.40)
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HClO(aq) + H+(aq) + 2 Cr2+(aq) = 2 Cr3+(aq) + Cl^-(aq) + H2O(l)	e^o = 1.80 V

E^o = 1.80 V – 0.40 V = 1.40 V

Eo = |__1.40 V____

The Cr³⁺ + e^- = Cr²⁺ half-reaction that is not given in your text table can be calculated using "skip potentials", a type of short-hand for the DG calculations ( = -nFe), ignoring the constant F

3(Cr³⁺ + 3 e^-= Cr) = 2(Cr²⁺ + 2 e^-= Cr) + 1(Cr³⁺ + e^- = Cr²⁺)

3(-0.74 V) = 2(-0.91 V) + e^o ; e^o = -2.22 V – (-1.82 V) = -0.40V