Problem #37Exercises on Galvanic Cells[Solved]

Questions: Using data in the Table of Standard Electrode Potentials shown in a previous problem, devise and represent Galvanic Cells made of the following redox couples, and calculate their EMF: (a)    (Fe²⁺, Fe) and (Ag⁺, Ag);  (b)   (Cl₂, 2Cl^-) and (Cu²⁺, Cu)

Solution:

(a) Methodology to be followed: (Fe²⁺, Fe) and (Ag⁺, Ag)

Step 1:

- Look at the electrode reduction potentials of the two redox couples and choose the couple with the higher reduction electrode potential, i.e. the easier to be reduced; this will be the reduction reaction at the cathode: (Ag⁺/Ag : 0.80 V)

- The remaining Couple will be the anode, where the oxidation reaction takes place: (Fe²⁺/Fe: -0.44 V)

- Then write the half-reactions:

Cathode/reduction: Ag⁺  +  e →  Ag       0.80V

Anode/oxidation:    Fe →  Fe²⁺  + 2e    +0.44 V   the sign changes because the reaction to the anode changes to oxidation.

Step 2:

- Equalize the number of electrons in both half-reactions if different:

Cathode/reduction: 2Ag⁺  + 2e  →  Ag 0.80V   the electrode potential remains the same

Anode/oxidation:    Fe →  Fe²⁺  + 2e        +0.44 V

Overall reaction: 2Ag⁺  + Fe  →  Ag +  Fe²⁺  1.24 V

This Galvanic cell is represented, starting by the anode at the left, as:

Fe/Fe²⁺//Ag⁺/Ag,  and its EMF = 1.24V

In a Galvanic Cell, the Anode where the oxidation reaction takes place constitutes the negative pole of the cell (source of electrons), whereas the Cathode, where the reduction reaction takes place constitutes the positive pole (see Figure in previous Problem 36).

(b) For (Cl₂, Cl^-) and (Cu²⁺, Cu), the same methodology is applied, and we get:

Cathode/reduction: Cl₂  + 2e →  2Cl^-       1.36 V

Anode/oxidation:    Cu →  Cu²⁺  + 2e    -0.34 V

Overall reaction:    Cl₂  + Cu  →  2Cl^-  +  Cu²⁺    1.02 V

This Galvanic cell is represented as: Cu/Cu²⁺//Cl₂/2Cl^-, and its EMF = 1.02 V

N.B: For a Galvanic cell to produce electricity, its EMF must be greater than zero volt (EMF > 0 V)