1. For a galvanic cell consisting of an iron electrode in contact with a solution of Fe+ ²(aq )  ions, and an aluminum electrode in contact with a solution of A+l ³(aq )  ion   s:

 

1. Determine the anode and cathode and write the half reactions
2. Calculate the standard cell potential
3. Write the shorthand cell notation

 

1. An electrochemical cell is constructed using electrodes based on the following half reactions:

Pb•<² aql  + 2e· -   Pb<sl               Au•aql + 3e· -   Au<sl

 

1. Which is the anode, and which is the cathode in the cell?
2. What is the standard cell potential?

 

1.- a) The solid iron electrode Fe(s) is the cathode (highest value of Eº).

Half reaction that occurs in the cathode : Fe²⁺(aq) + 2e^- ----> Fe(s) (reduction half reaction)

.- The solid Aluminum electrode Al (s) is the anode (lowest value of Eº)

Half reaction that occurs in the cathode : Al(s) ----> Al³⁺(aq) + 3e^- (oxidation half reaction).

b) Eº_cell = + 1.21 V.

c) Al (s) | Al³⁺(aq) || Fe²⁺(aq) | Fe (s)

 

2.- a).- The solid gold electrode Au(s) is the cathode (highest value of Eº).

Half reaction that occurs in the cathode : Au³⁺(aq) + 3e^- ----> Au(s)

.- The solid lead electrode Pb (s) is the anode (lowest value of Eº)

Half reaction that occurs in the anode : Pb²⁺(aq) + 2e^- ----> Pb (s)

 

b).- Eº_cell = + 1.63 V.

Step-by-step explanation

1.- For a galvanic cell consisting of an iron electrode in contact with a solution of Fe²⁺(aq) ions, and an aluminum electrode in contact with a solution of Al³⁺ (aq ) ions:

 

a .- Determine the anode and cathode and write the half reactions

In a galvanic cell, to know which electrode is the cathode and which is the anode, the standard reduction potential Eº_red must be evaluated. This value are tabulated and can be found in any chemical bibliography.

The standard reduction potential is the reference voltage that generates a half-reaction when the process of gaining electrons (reduction) occurs. The larger the standard reduction potential value, the easier the electrode will be reduced. The electrode that has the greatest standard reduction potential when building a galvanic cell corresponds to the cathode, and the one with the lowest Eº_red., the anode.

From standard reduction potential Table (In this case taked from Chang 7th edition)

Al³⁺(aq) + 3e^- -----> Al(s) Eº _{Al3+/ Alº} = -1.66 V.

Fe²⁺(aq) + 2e^- ----> Fe(s) Eº _{Fe2+/ Feº} = -0.45 V

 

Among the two reduction potential values, that of iron is the highest at -0.45 volts, relative to the standard reduction potential of aluminum which is -1.66 V.

 

The solid iron electrode Fe(s) is the cathode (highest value of Eº).

Half reaction that occurs in the cathode : Fe²⁺(aq) + 2e^- ----> Fe(s) (reduction half reaction)

 

The solid Aluminum electrode Al (s) is the anode (lowest value of Eº)

Half reaction that occurs in the cathode : Al(s) ----> Al³⁺(aq) + 3e^- (oxidation half reaction)

 

b.- Calculate the standard cell potential

Using the standard potential equation:

Eº_cell = Eº reduction cathode - Eº reduction Anode

Eº_cell = Eº _{Fe2+/ Feº} - Eº _{Al3+/ Alº}

Eº_cell = - 0.45 V - (-1.66 V) = 1.21 V

Eº_cell = + 1.21 V.

 

.- Write the shorthand cell notation:

By convention, the species of the cell that acts as the anode (oxidation) are written on the left.

On the right are written the species of the cell that acts as a cathode (reduction).

 

In general, the abbreviated notation is written from left to right as follows:

Anode | Anodic solution || Cathodic dissolution | Cathode

A vertical line indicates a phase change. A double vertical line indicates a porous membrane or salt bridge.

 

For this reaction:

Al (s) | Al³⁺(aq) || Fe²⁺(aq) | Fe (s)

 

2.- An electrochemical cell is constructed using electrodes based on the following half reactions:

Pb⁺² (aq)  + 2e^- ------->  Pb(s)             Au³⁺ (aq) + 3e^-  ------->  Au (s)

 

.- Which is the anode, and which is the cathode in the cell?

Using the tabulated value of standard reduction potential for the half cell, we can determine the cathode and the anode.

Pb⁺² (aq)  + 2e^- ------->  Pb(s) Eº= -0.13 V

Au³⁺ (aq) + 3e^-  ------->  Au (s) Eº = +1.50 V.

 

The solid gold electrode Au(s) is the cathode (highest value of Eº).

Half reaction that occurs in the cathode : Au³⁺(aq) + 3e^- ----> Au(s)

 

The solid lead electrode Pb (s) is the anode (lowest value of Eº)

Half reaction that occurs in the anode : Pb²⁺(aq) + 2e^- ----> Pb (s)

 

.- What is the standard cell potential?

Using the standard potential equation:

Eº_cell = Eº reduction cathode - Eº reduction Anode

Eº_cell = Eº _{Au3+/ Auº} - Eº _{Pb2+/ Pbº}

Eº_cell = +1.50 V - (-0.13 V) = +1.63 V

Eº_cell = + 1.63 V.