Everything Science Grade 12

CHAPTER 4. ELECTROCHEMICAL REACTIONS 4.4

Example 4: Calculating the emf ofa cell

QUESTION

The following reaction takes place:

Cu(s) + Ag+(aq)→ Cu2+(aq) + Ag(s)

1. Represent the cell using standard notation.


2. Calculate the cell potential (emf) of the electrochemical cell.

SOLUTION

Step 1 : Write equations for thetwo half reactions involved

Cu2++ 2e−� Cu (E◦V = 0. 16 V)

Ag++ e−� Ag (E◦V = 0. 80 V)

Step 2 : Determine which reaction takes place at the cathode and which

is the anode reaction

Both half-reactions havepositive electrode potentials, but the sil-

ver half-reaction has a higher positive value. In other words, silver

does not form ions easily, and this must be the reduction half-

reaction. Copper is theoxidation half-reaction. Copper is oxi-

dised, therefore this is the anode reaction. Silver is reduced and

so this is the cathode reaction.

Step 3 : Represent the cell using standard notation

Cu|Cu2+(1mol· dm−^3 )||Ag+(1mol· dm−^3 )|Ag

Step 4 : Calculate the cell potential

E◦(cell)= E◦(cathode) - E◦(anode)

= +0. 80 − (+0.34)

= +0. 46 V