Chapter 20, Problem 42A

Interpretation Introduction

(a)

Interpretation:

The anode is to be identified.

Concept introduction:

A galvanic cell is represented by writing the anode where oxidation occurs on the left hand side and cathode where reduction occurs on right hand side.

Answer to Problem 42A

The zinc electrode is the anode.

Explanation of Solution

The zinc electrode where electrons are released or oxidation occurs is called anode. So, the zinc electrode is the anode.

(b)

Interpretation Introduction

Interpretation:

The cathode electrode is to be identified.

Concept introduction:

A galvanic cell is represented by writing the anode where oxidation occurs on the left hand side and cathode where reduction occurs on right hand side.

Answer to Problem 42A

The silver electrode is the cathode.

Explanation of Solution

The silver electrode where electrons are accepted or reduction occurs is called cathode. So, silver electrode is the cathode.

(c)

Interpretation Introduction

Interpretation:

The electrode where oxidation is occurs is to be explained.

Concept introduction:

Oxidation is the process in which loss of electron is occurring.

Answer to Problem 42A

Oxidation occurs at the zinc electrode or at the anode.

Explanation of Solution

The zinc electrode where electrons are released or oxidation occurs is called anode. So, the zinc electrode is the anode

(d)

Interpretation Introduction

Interpretation:

The electrode where reduction occurs is to be identified..

Concept introduction:

Reduction is the process in which gain of electrons are occurs.

Answer to Problem 42A

Reduction is occurs at the silver electrode or at the cathode.

Explanation of Solution

The silver electrode where electrons are accepted or reduction occurs is called cathode. So, silver electrode is the cathode.

Interpretation Introduction

(e)

Interpretation:

The current flow in which direction is to be identified.

Concept introduction:

Reduction is the process in which gain of electrons are occurs

Oxidation is the process in which loss of electron is occurring

Answer to Problem 42A

The current flow form positive terminal or cathode to negative terminal or anode. So, the electric current flows from silver electrode to zinc electrode.

Explanation of Solution

The zinc electrode released electrons while silver electrode accepted the electrons. There is flow of electrons from negative terminal to positive terminal. As a result of which, the electric current is flow from positive or silver electrode to negative or zinc electrode.

Interpretation Introduction

(f)

Interpretation:

The direction of positive ion is to be identified.

Concept introduction:

The direction of positive ions is opposite to the direction of flow of electrons.

Answer to Problem 42A

The positive ions are flow from positive terminal or cathode to negative terminal or anode.

Explanation of Solution

The flow of electron from one electrode to other in one direction, motion of ionic species through the salt bridge carries negative charge through the solution in opposite direction. So, the positive ions are flow from cathode to anode.

Interpretation Introduction

(h)

Interpretation:

The cell potential is to be determined.

Concept introduction:

The cell potential is the difference between the cell potential of cathode and electrode.

The formula is given as

$\begin{array}{l}{\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{where,}{\text{E°}}_{\text{cell}}\text{=}\text{Standard}\text{Cell}\text{potential}\\ {\text{E°}}_{\text{cathode}}\text{=Standard}\text{reduction}\text{potential}\text{at}\text{cathode}\\ {\text{E°}}_{\text{anode}}\text{=Standard}\text{reduction}\text{potential}\text{at}\text{anode}\\ \end{array}$

Answer to Problem 42A

According to the given information, the cell potential is given as

$\begin{array}{l}{\text{E°}}_{\text{cell}}{\text{= E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=0}\text{.7996-}\left(\text{-0}\text{.7618}\right)\\ \text{=1}\text{.564}\text{V}\end{array}$

The cell potential is $\text{1}\text{.564}\text{V}$

Explanation of Solution

The cell potential is the difference between the cell potential of cathode and electrode.

The formula is given as

$\begin{array}{l}{\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{where,}{\text{E°}}_{\text{cell}}\text{=}\text{Standard}\text{Cell}\text{potential}\\ {\text{E°}}_{\text{cathode}}\text{=Standard}\text{reduction}\text{potential}\text{at}\text{cathode}\\ {\text{E°}}_{\text{anode}}\text{=Standard}\text{reduction}\text{potential}\text{at}\text{anode}\\ \end{array}$

According to the given information, the cell potential is given as

$\begin{array}{l}{\text{E°}}_{\text{cell}}{\text{= E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=0}\text{.7996-}\left(\text{-0}\text{.7618}\right)\\ \text{=1}\text{.564}\text{V}\end{array}$

The cell potential is $\text{1}\text{.564}\text{V}$