Chapter 18, Problem 18.66QE

Interpretation Introduction

Interpretation:

Ratio of $\left[{\text{Cd}}^{\text{+2}}\right]/{\left[{\text{Ag}}^{\text{+}}\right]}^2$ for the given voltaic cell which has a voltage of $\text{0}\text{.425 V}$ has to be determined.

  $\text{ 2}{\text{Ag}}^{\text{+}}\text{(aq) +Cd}\text{(s) }\to \text{ 2}\text{Ag}{\text{(s) + Cd}}^{+2}\text{(aq)}$

Concept introduction:

Electrochemical cells:

In all electrochemical cells, oxidation occurs at anode and reduction occurs at cathode.

An anode is indicated by negative sign and cathode is indicated by the positive sign.

Electrons flow in the external circuit from the anode to the cathode.

In the electrochemical cells two half cells are connected with salt bridge. It allows the cations and anions to move between the two half cells.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

  ${\text{E}}_{\text{cell}}^{\text{o}}{\text{=E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}$

If ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, then the reaction is predicted to be spontaneous at that direction.

If ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, then the reaction is predicted to be spontaneous at reverse direction.

The relation between standard cell potential and equilibrium constant is as follows.

  $\text{log}\text{K = }\frac{{\text{nE}}^{\text{0}}}{\text{0}\text{.0591}}\text{ at 298K}$