Chapter 15, Problem 15.6P

a)

Interpretation Introduction

Interpretation:

When a cell was prepared by dipping a $\text{Cu}$ wire and a saturated calomel electrode into $\text{0}\text{.10}\text{M}$ ${\text{CuSO}}_{\text{4}}$ solution. The $\text{Cu}$ wire was connected to positive terminal and the calomel electrode was connected to the negative terminal

The half-reaction for the $\text{Cu}$ metal has to be written.

Concept Introduction:

Half-Reaction:

It is a component of a redox reaction. Which can be either oxidation or reduction reaction component. It is considered by the change in oxidation states of individual substances.

b)

Interpretation Introduction

Interpretation:

When a cell was prepared by dipping a $\text{Cu}$ wire and a saturated calomel electrode into $\text{0}\text{.10}\text{M}$ ${\text{CuSO}}_{\text{4}}$ solution. The $\text{Cu}$ wire was connected to positive terminal and the calomel electrode was connected to the negative terminal

The Nernst equation for the $\text{Cu}$ metal has to be written.

Concept Introduction:

Nernst equation:

In electrochemistry, the standard reduction potentials are related with the temperature and gas constant and standard electrode potentials and concentrations.

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{0}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{nF}}\text{log}\frac{{\left[\text{cathode}\right]}^{\text{n}}}{{\left[\text{anode}\right]}^{\text{n}}}$

c)

Interpretation Introduction

Interpretation:

When a cell was prepared by dipping a $\text{Cu}$ wire and a saturated calomel electrode into $\text{0}\text{.10}\text{M}$ ${\text{CuSO}}_{\text{4}}$ solution. The $\text{Cu}$ wire was connected to positive terminal and the calomel electrode was connected to the negative terminal

The cell voltage of $\text{Cu}$ metal has to be calculated.

Concept Introduction:-

Cell Voltage:

Cell voltage is the difference between cathode potential and anode potential.

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{+}}\text{-}{\text{E}}_{\text{-}}$