Chapter 20.1, Problem 1PP

Interpretation Introduction

Interpretation:

The given cell reaction is to be balanced and the cell potential is to be determined.

Concept introduction:

An electrochemical reaction is balanced by balancing the oxidation number on both sides of the reaction. The difference between the electrode potential of the two electrodes consisting of an electrochemical cell is known as cell potential.

Answer to Problem 1PP

The cell potential is $\text{1}\text{.32}\text{V}$ .

The $\text{Sn}$ electrode is an anode where oxidation occurs and $\text{Pt}$ electrode is a cathode where reduction occurs.

Explanation of Solution

The overall reaction of the given pair of half cells reaction is written as

  $\begin{array}{l}\text{Sn}\left(\text{s}\right){\text{+Pt}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{Sn}}^{\text{+2}}\left(\text{aq}\right)\text{+Pt}\left(\text{s}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ \text{Sn}|{\text{Sn}}^{\text{+2}}\Vert {\text{Pt}}^{\text{+2}}|\text{Pt}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=1}\text{.18-}\left(\text{-0}\text{.1375}\right)\\ \text{=1}\text{.32}\text{V}\end{array}$

The cell potential is $\text{1}\text{.32}\text{V}$

The $\text{Sn}$ electrode is an anode where oxidation occurs and $\text{Pt}$ electrode is a cathode where reduction occurs.

Conclusion

Thus, the cell potential is $\text{1}\text{.32}\text{V}$ .

The $\text{Sn}$ electrode is an anode where oxidation occurs and $\text{Pt}$ electrode is a cathode where reduction occurs.