Chapter 20, Problem 66A

Interpretation Introduction

Interpretation:

The oxidation and reduction half reactions are to be written, anode and cathode are to be identified and the electron flow direction is to be mentioned.

Concept introduction:

In a voltaic cell, the overall cell potential is positive, oxidation occurs at the anode and reduction occurs at the cathode. The electrons flow from the anode to the cathode.

Answer to Problem 66A

The oxidation half cell reaction:

$Cr(s)\to C{r}^{3+}(aq)+3e^{-}$

The reduction half cell reaction:

$A{g}^{+}(aq)+e^{-}\to Ag(s)$

The chromium electrode is the anode and the silver electrode is the cathode.

Electrons flow from the chromium electrode to the silver electrode.

Explanation of Solution

In the above cell, the reduction potential of Ag is higher than that of chromium. Therefore for the cell potential to be positive, reduction will occur at silver electrode and oxidation will occur at the chromium electrode which makes them cathode and anode respectively. And elctrons always flow from anode to cathode which determines the electorn flow.

Conclusion

The above equations and conclusions have been determined from the reduction potentials of the respective half cells.