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CHEMISTRY 104 – WORKSHEET #12
Ch. 20: Electrochemistry (Part I)
http://www.chem.wisc.edu/areas/clc (Resource page)
Prepared by Dr. Tony Jacob
ASSIGN OXIDATION NUMBERS
Elements = 0; In a compound: Group 1A = +1, Group 2A = 2, F = -1, H = +1 (usually), O = -2 (usually)
Sum Rule: Sum of all the oxidation numbers = total charge on compound
oxidation
0
-
+
oxidation number numberline
reduction
Is it a redox reaction? Look for change in oxidation numbers from reactants to products
LEO the lion goes GER (LEO - Lose electrons is oxidation; GER - Gain electrons is reduction)
Reducing agent - causes a chemical to be reduced while it is oxidized in the process
Oxidizing agent - causes a chemical to be oxidized while it is reduced in the process
BALANCING REDOX REACTIONS - Half-Reaction Method
1. Separate into two half-reactions.
2. Balance redox atom.
3. Add H2O to balance O.
4. Add H+ to balance H.
5. Add e- to balance the total charges.
6. Multiply half-reactions so #e- are the same in each half-reaction.
7. Add half-reactions; cancel e- and H2O if possible.
8. If basic soln: Add same #OH- to both sides of rxn; #OH- = #H+; H++OH- → H2O; cancel H2O if possible.
ELECTROCHEMICAL CELLS:
Voltaic or Galvanic cell: spontaneous reaction (i.e., a battery); Ecell > 0 → product favored
Electrolytic cell: non-spontaneous reaction; energy needed for reaction; Ecell < 0 → reactant favored
Drawing an electrochemical cell: label electrodes and solution; label which cell is the anode/oxidation and
cathode/reduction; draw direction/flow of electrons in wire (always anode → cathode) and anions/cations in
salt bridge; determine Eocell; determine what if anything ppt/dissolves; write half-rxns under each cell;
understand purpose of inert electrode if one is present
Image by Quinn J. Mattsson
Zn(s) | Zn+2(aq) (1M) || Cu+2(aq) (1M) | Cu(s)
SHE = standard hydrogen electrode: reference cell; Eo = 0.00V; 2H+ + 2e- → H2 (1M, 1atm, 25˚C)
UNITS: V = J/C; V = volts; C = coulombs (unit of charge); V = volts
HOW E CHANGES: 1. Reverse rxn: E → –E; 2. Multiply rxn by c: no change; 3. Add rxns → add E’s
LINE NOTATION: anode electrode(s) | anode solution(aq) || cathode solution(aq) | cathode electrode(s)
the single “|” means phase change: aqueous to solid; the double “||” means: separation between half-cells
Ex: Cu(s) | Cu+2(aq) (1M) || Fe+3(aq) (1M), Fe+2(aq) (1M) | Pt(s)
INERT ELECTRODES: If half-reaction has no metal, e.g., Fe+3(aq) + e- → Fe+2(aq), an inert electrode is
used to conduct e-; common inert electrodes: Pt(s), Au(s), Cgraphite(s)
FIND Eocell given 2 half-reactions
• given 2 reduction half-reactions, one reaction must be reversed to yield an oxidation reaction; change sign
of Eo and add to other reaction so Eocell is positive for a spontaneous reaction – or –
•
Eocell = Eocathode – Eoanode
STRONGER OXIDIZING /REDUCING AGENT: use
Standard Reduction Table
The more positive the E → the more spontaneous
(Note: There is a table of standard reduction potentials at the back of this handout.)
1. The oxidation number of sulfur in calcium thiosulfate, CaS2O3, is
a. -2
b. 0
c. +2
d. +4
e. +6
2. Use the following balanced reaction to answer the next 3 questions.
8S2-(aq) + 16NO3-(aq) + 32H+(aq) → 16NO2(g) + S8(s) + 16H2O(l)
I. What chemical species is oxidized?
a. S-2
b. NO3-
c. NO2
d. S8
e. H+
II. What chemical species is the oxidizing agent?
a. S-2
b. NO3c. NO2
d. S8
e. H+
III. How many electrons were transferred in the reaction?
a. 1
b. 2
c. 8
d. 16
e. 32
3. Balance the redox reaction below under acidic conditions. In the overall balanced reaction the coefficient in
front of H2O is
a. 1
b. 2
c. 3
d. 4
e. 5
-2
+3
Cr2O7 (aq) + NO(g) → Cr (aq) + NO3 (aq)
4. In the list below, select the strongest reducing agent using the table of standard reduction potentials at the
back of this handout.
a. Ni
b. Fe+2
c. PbSO4
d. Fe
e. Au
5. Using the standard reduction table, which of the metals will not spontaneously reduce H+(aq) to H2(g) under
standard conditions?
a. Ag(s)
b. Sn(s)
c. Zn(s)
d. Na(s)
6. Given the balanced reaction below, what is the line notation of the electrochemical cell under standard
conditions?
2Fe+3(aq) + H2(g) → 2Fe+2(aq) + 2H+(aq)
a. Pt(s) | Fe+2(aq) (1.0M), Fe+3(aq) (1.0M) || H+(aq) (1.0M) | H2(g) (1.0atm) | Pt(s)
b. Pt(s), H2(g) (1.0atm), H+(aq) (1.0M) || Fe+2(aq) (1.0M), Fe+3(aq) (1.0M), Pt(s)
c. H2(g) (1.0atm) | H+(aq) (1.0M) || Fe+2(aq) (1.0M), Fe+3(aq) (1.0M)
d. Pt(s) | H2(g) (1.0atm) | H+(aq) (1.0M) || Fe+2(aq) (1.0M), Fe+3(aq) (1.0M) | Pt(s)
e. H2(g) (1.0atm) | 2H+(aq) (1.0M) || 2Fe+2(aq) (1.0M), 2Fe+3(aq) (1.0M)
7. The half-reactions below are part of a galvanic cell with a KBr salt bridge. Answer the questions below.
Ag+(aq) + e- → Ag(s)
Eo = 0.80V
Sn+2(aq) + 2e- → Sn(s)
Eo = -0.14V
I. What is the value of Eocell?
a. 0.66V
b. 0.94V
c.1.46V
d. 1.74V
e. -0.94V
II. The electrons
a. will travel from the Sn(s) through the wire to Ag(s)
b. will travel from the Ag(s) through the wire to Sn(s)
c. will travel from the Sn+2(aq) solution through the salt bridge to the Ag+(aq) solution
d. will travel from the Ag+(aq) solution through the salt bridge to the Sn+2(aq) solution
e. will not travel through the system since this is a non-spontaneous electrochemical cell
III. What will occur at the anode?
a. the Sn+2(aq) will be reduced to Sn(s)
b. the Ag+(aq) will be reduced to Ag(s)
c. the Sn(s) will be oxidized to Sn+2(aq)
d. the Ag(s) will be oxidized to Ag+(aq)
e. the Sn+2(aq) will be plated out onto the Ag(s)
IV. Which statement about the salt bridge is correct? If statements “a”-“d” are incorrect, select answer “e”.
a. Only Br- will travel to the Sn/Sn+2 half-cell.
b. K+ will travel to the Sn/Sn+2 half-cell while the Br- will travel to the Ag/Ag+ half-cell.
c. Br- will travel to the Sn/Sn+2 half-cell while the K+ will travel to the Ag/Ag+ half-cell.
d. Only e- travel through the salt bridge.
e. All of the above statements are incorrect.
8. Given the line notation for the electrochemical cell, what is the balanced chemical reaction?
Pt(s) | Sn+2(aq) (1.0M), Sn+4(aq) (1.0M) || Fe+3(aq) (1.0M) | Fe(s)
a. 3Fe+3(aq) + 2Sn+2(aq) → 3Fe(s) + 2Sn+4(aq)
b. Fe+3(aq) + Sn+2(aq) → Fe(s) + Sn+4(aq)
c. 2Fe(s) + 3Sn+4(aq) → 2Fe+3(aq) + 3Sn+2(aq)
d. 2Fe+3(aq) + 3Sn+2(aq) → 2Fe(s) + 3Sn+4(aq)
e. 2Fe+3(aq) + 3Sn+2(aq) + Pt(s) + 4Cl-(aq) → 2Fe(s) + 3Sn+4(aq) + PtCl4-2(aq)
9. The following electrochemical cell line notation and potential are given below.
Ni(s) | Ni+2(aq) (0.025M) || Cr+3(aq) (1.25M) | Cr(s)
Ecell = -0.44V
Which statement is correct?
a. The reaction is spontaneous.
b. The value of K is greater than 1.
c. The value of ΔG > 0.
d. This is a voltaic cell.
e. This reaction is run under standard conditions.
ANSWERS
1. c {Ca has a charge of +2; so S2O3 has a charge of –2 → S2O3-2; assign O = -2 and from sum rule S = +2;
2.
Sum rule: -2 = 3(O) + 2(S) = 3(-2) + 2(S); solve for S → +2}
I. a {S-2: S = -2; S8: S = 0; -2 → 0 → oxidized}
II. b {NO3-: N = +5; NO2: N = +4; +5 → +4 → reduced = oxidizing agent}
III. d {S: each S goes from -2 → 0 and loses 2e-; 2e-/S x 8S = 16e-; or using N: each N goes from +5 → +4 and gains 1e-;
1e-/N x 16N = 16e-}
{bal rxn: Cr2O7-2 + 2NO + 6H+ → 2Cr+3 + 3H2O + 2NO3-; both #atoms and charge must be balanced in a reaction}
{to be a reducing agent the chemical must be oxidized; so the reactions are: Ni → Ni+2 + 2e-, Eo = 0.25;
3. c
4. d
Fe+2 → Fe+3 + e-, Eo = -0.771; PbSO4 + 2H2O → PbO2 + SO4-2 + 4H2O + 2e-, Eo = -1.685; Fe → Fe+2 + 2e-, Eo = 0.44;
Au → Au+3 + 3e-, Eo = -1.50; the reaction furthest down the table on the right side is the strongest reducing agent → Fe}
{2H+(aq) + 2e- → H2(g); Eo = 0V; Ag(s) → Ag+(aq) + e-; Eo = -0.80V; 2H+(aq) + 2Ag(s) → H2(g) + 2Ag+(aq);
Eocell = -0.80V; since Eocell < 0 (-) → non-spontaneous; the other metals will yield a positive Eocell and are therefore
spontaneous; e.g., 2H+(aq) + 2e- → H2(g); Eo = 0V; Sn(s) → Sn+2(aq) + 2e-; Eo = 0.14V;
2H+(aq) + Sn(s) → Sn+2(aq) + H2(g); Eocell = 0.14V; etc.}
d {left side of line notation = anode/oxidation: H2(g) → 2H+(aq) + 2e-; right side is cathode/reduction:
Fe+3(aq) + e- → Fe+2(aq); Pt(s) is an inert electrode and needs to be included when there is no metal electrode as part of the half-
5. a.
6.
7.
reaction which is the case for both half-reactions; a “|” is used between phases (e.g., between (aq) and (s)) and “||” used as the
divider between the half-cells; stoichiometric coeffieients are not included}
I. b {galvanic cell → spontaneous reaction → Eocell > 0 (+); reverse the Sn reaction: Sn(s) → Sn+2(aq) + 2e-, Eo = +0.14V;
multiply the silver reaction by 2 so electrons will cancel: 2Ag+(aq) + 2e- → 2Ag(s); Eo = +0.80V; add half reactions:
Sn(s) + 2Ag+(aq) → Sn+2(aq) + 2Ag(s) and add Eo: Eocell = 0.80 + 0.14 = 0.94V}
II. a {electrons only travel through the wire and electrodes and never through the solution or salt bridge; electrons go from the
anode → cathode; anode (ox): Sn/Sn+2; cathode (red): Ag/Ag+}
c {anode = ox; Sn is oxidized to Sn+2: Sn(s) → Sn+2(aq) + 2e-}
III.
IV. c {anions travel in the opposite way of electrons: cathode → anode; the anode is Sn/Sn+2 so Br- will go to Sn/Sn+2;
8.
at the same time, K+ can travel to the cathode though this is less dominant}
d {left side of line notation = anode/oxidation; Pt(s) is an inert electrode and should not be part of the overall reaction; electrons
needs to be balanced: Sn+2(aq) → Sn+4(aq) + 2e- and Fe+3(aq) + 3e- → Fe(s) so Sn+2 reaction should be multiplied by 3 and
9.
Fe+3 reaction multiplied by 2}
c {since Ecell < 0 (-) → nonspontaneous reaction; nonspontaneous reactions have K < 1; voltaic cells are spontaneous; since the
[Ni+2] ≠ 1M nor does [Cr+3] ≠ 1M → nonstandard conditions; ΔG < 0 (-) for spontaneous reactions and ΔG > 0 (+) for
nonspontaneous reactions}
Standard Reduction Potentials
F2(g) + 2e- → 2F-(aq)
Co+3(aq) + e- → Co+2(aq)
+2.87V
PbO2(s) + HSO4-(aq) + 3H+(aq) + 2e- → PbSO4(s) + 2H2O(l)
MnO4-(g) + 8H+(aq) + 5e- → Mn+2(aq) + 4H2O(l)
Au+3(aq) + 3e- → Au(s)
+1.685V
2ClO3-(aq) + 12H+(aq) + 10e- → Cl2(g) + 6H2O(l)
+1.47V
Cl2(g) + 2e- → 2Cl-(aq)
+1.36V
Cr2O7-2(g) + 14H+(aq) + 6e- → 2Cr+3(aq) + 7H2O(l)
+1.33V
O2(g) + 4H+(aq) + 4e- → 2H2O(l)
Br2(g) + 2e- → 2Br-(aq)
+1.23V
NO3-(aq) + 4H+(aq) + 3e- → NO(g) + 2H2O(l)
+0.96V
Ag+(aq) + e- → Ag(s)
+0.80V
Fe+3(aq) + e- → Fe+2(aq)
+0.77V
O2(g) + 2H+(aq) + 2e- → H2O2(aq)
+0.68V
MnO4-(aq) + 2H2O(l) + 3e- → MnO2(s) + 4OH-(aq)
I2(g) + 2e- → 2I-(aq)
+0.59V
Cu+(aq) + e- → Cu(s)
+0.521V
O2(g) + 2H2O(l) + 4e- → 4OH-(aq)
+0.40V
Cu+2(aq) + 2e- → Cu(s)
+0.34V
AgCl(s) + e- → Ag(s) + Cl-(aq)
Sn+4(aq) + 2e- → Sn+2(aq)
+0.222V
Cu+2(aq) + e- → Cu+(aq)
AgBr(s) + e- → Ag(s) + Br-(aq)
+0.153V
+1.842V
+1.51V
+1.50V
+1.06V
+0.54V
+0.154V
+0.095V
2H+(aq) + 2e- → H2(g)
0.00V
Pb+2(aq) + 2e- → Pb(s)
-0.126V
Sn+2(aq) + 2e- → Sn(s)
CuI(s) + e- → Cu(s) + I-(aq)
-0.136V
Co+2(aq) + 2e- → Co(s)
Ni+2(aq) + 2e- → Ni(s)
-0.277V
PbSO4(s) + H+(aq) + 2e- → Pb(s) + HSO4-(aq)
-0.356V
Cd+2(aq) + 2e- → Cd(s)
-0.40V
Cr+3(aq) + e- → Cr+2(aq)
Fe+2(aq) + 2e- → Fe(s)
-0.41V
Cr+3(aq) + 3e- → Cr(s)
Zn+2(aq) + 2e- → Zn(s)
-0.74V
2H2O(l) + 2e- → H2(g) + 2OH-(aq)
-0.83V
Mn+2(aq) + 2e- → Mn(s)
-1.18V
Al+3(aq) + 3e- → Al(s)
Mg+2(aq) + 2e- → Mg(s)
-1.66V
Na+(aq) + e- → Na(s)
K+(aq) + e- → K(s)
-2.71V
Li+(aq) + e- → Li(s)
-3.05V
-0.185V
-0.28V
-0.44V
-0.76V
-2.37V
-2.925V