S17E3 - Galvanic Cell Examples and Balancing Half-Reactions

Today we're going to discuss three galvanic cell problems.

FYI:  galvanic cell  =  electrochemical cell  =  voltaic cell

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Galvanic Cell Example #1

ex:  A galvanic cell is based on the following unbalanced equation:

MnO₄^-(aq)  +  8H⁺  +  ClO₃^-(aq)  →  ClO₄^-(aq)  +  Mn²⁺(aq)  +  H₂O(l)

Give the balanced half-reactions, the balanced cell reaction, calculate E°cell, and give the cell's line notation.
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answer:

➞  half-reactions - use the Table of Standard Reduction Potentials from my previous video post to get:

(anode) oxidation,
ClO₃^-  +  H₂O  →  ClO₄^-  +  2H⁺  +  2e^-  , E°_ox = -1.19 V

(cathode) reduction,
MnO₄^-  +  5e^-  +  8H⁺ →  Mn²⁺  +  4H₂O   ,  E°_red = +1.51 V

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➞  Then multiply each reaction through to get the number of electrons (10e^-) to cancel:

2MnO₄^-  +  10e^-  +  16H⁺ →  2Mn²⁺  +  8H₂O   , E°_red = +1.51 V
5ClO₃^-  +  5H₂O  →  5ClO₄^-  +  10H⁺  +  10e^-   ,  E°_ox = -1.19 V

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➞  Eliminate and reduce to get the balanced cell reaction:

2MnO₄^-  +  5ClO₃^-  +  6H⁺ →  2Mn²⁺  +  3H₂O  +  5ClO₄^- 
E°_cell  =  +0.32 V

See here...

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➞  Finally, we can write the line notation:

Pt(s) | ClO₃^-(aq), ClO₄^-(aq), H⁺(aq) || H⁺(aq), MnO₄^-(aq), Mn²⁺(aq) | Pt(s)

or:

What's "Pt" ? -- Platinum (Pt) is an inert metal that must be used as the conducting electrode when there's no metal(s) involved in the anode or cathode.

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Galvanic Cell Example #2

ex:  "Completely describe" the following galvanic cell, given just its half-reactions:

Fe²⁺  +  2e^-→  Fe(s)   ,  E°_red  =  -0.44 V
MnO₄^-  +  5e^-  +  8H⁺  →  Mn²⁺  +  4H₂O  , E°_red  =  +1.51 V
_________
answer:

When "completely describing" a galvanic cell, we need to know how to:

➞ 1. write the overall cell reaction (E°_cell > 0),

➞ 2. calculate E°_cell,

➞ 3. designote anode and cathode,

➞ 4. show direction of electron flow and report ions in each compartment (line notation).

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Ok, looking at the two reduction half-reactions, the E°_red  =  +1.51 V is probably written in the correct direction because we want E°_cell to be overall positive (+ volts).

So we'll reverse the Fe / Fe²⁺ half-reaction...

(anode) oxidation,
5 x  [ Fe(s)  →  Fe²⁺  +  2e^- ]   ,  E°_ox  =  +0.44 V

(cathode) reduction,
2 x  [ MnO₄^-  + 5e^-  + 8H⁺ →  Mn²⁺  + 4H₂O ]  , E°_red  =  +1.51 V

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➞  Now we can write the line notation:

Fe(s) | Fe²⁺(aq) || H⁺(aq), MnO₄^-(aq), Mn²⁺(aq) | Pt(s)

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➞  Finally, we can sketch the galvanic cell, designate the anode / cathode, and determine the direction of electron flow.

In a galvanic cell, electrons flow from anode to cathode!! 

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Galvanic Cell Example #3

ex:  Given the following half-reactions, describe completely the galvanic cell:

Ag⁺  +  e^-  →  Ag(s)  , E° =  + 0.80 V

Fe³⁺  +  e^-  →  Fe²⁺  , E° =  + 0.77 V
_________
answer:

We want E°_cell > 0, so "flip" the 2nd half-reaction to get:

(cathode) reduction,
Ag⁺  +  e^-  →  Ag(s)   , E° =  + 0.80 V

(anode) oxidation,
Fe²⁺  →  Fe³⁺  +  e^-   , E° =  - 0.77 V

Adding these two half-reactions together gives the following:

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➞  Line notation:

                   Pt(s) | Fe²⁺, Fe³⁺ || Ag⁺ | Ag(s)
         anode                                                    cathode

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➞  Sketch of the galvanic cell:

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Coming up in my next post on SECTION 17 - Electrochemistry,

We'll discuss Concentration Cells and the Relationship between Free Energy (ΔG) and Cell Potential (E°cell).