Half reactions for silver-silver chloride along with calomel electrodes used as reference have to be written. Concept Introduction: Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode. For a general representation the oxidation half cell is drawn in left separated by double vertical parallel lines while reduction half cell comes at right. Expression to compute E cell is written as follows: E cell ° = E right − E left

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Answer 1

Question

Chapter 14, Problem 14.22P

(a)

Interpretation Introduction

Interpretation:

Half reactions for silver-silver chloride along with calomel electrodes used as reference have to be written.

Concept Introduction:

Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode. For a general representation the oxidation half cell is drawn in left separated by double vertical parallel lines while reduction half cell comes at right.

Expression to compute Ecell is written as follows:

Ecell°=Eright−Eleft

(b)

Interpretation Introduction

Interpretation:

Voltage of cell drawn has to be predicted.

Concept Introduction:

Refer to part (a).

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6. Consider the following exothermic reaction below. 2Cu2+(aq) +41 (aq)2Cul(s) + 12(aq) a. If Cul is added, there will be a shift left/shift right/no shift (circle one). b. If Cu2+ is added, there will be a shift left/shift right/no shift (circle one). c. If a solution of AgNO3 is added, there will be a shift left/shift right/no shift (circle one). d. If the solvent hexane (C6H14) is added, there will be a shift left/shift right/no shift (circle one). Hint: one of the reaction species is more soluble in hexane than in water. e. If the reaction is cooled, there will be a shift left/shift right/no shift (circle one). f. Which of the changes above will change the equilibrium constant, K?

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Answer 2

Question

Chapter 14, Problem 14.22P

(a)

Interpretation Introduction

Interpretation:

Half reactions for silver-silver chloride along with calomel electrodes used as reference have to be written.

Concept Introduction:

Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode. For a general representation the oxidation half cell is drawn in left separated by double vertical parallel lines while reduction half cell comes at right.

Expression to compute Ecell is written as follows:

Ecell°=Eright−Eleft

(b)

Interpretation Introduction

Interpretation:

Voltage of cell drawn has to be predicted.

Concept Introduction:

Refer to part (a).

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Answer 3

Question

Chapter 14, Problem 14.22P

(a)

Interpretation Introduction

Interpretation:

Half reactions for silver-silver chloride along with calomel electrodes used as reference have to be written.

Concept Introduction:

Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode. For a general representation the oxidation half cell is drawn in left separated by double vertical parallel lines while reduction half cell comes at right.

Expression to compute Ecell is written as follows:

Ecell°=Eright−Eleft

(b)

Interpretation Introduction

Interpretation:

Voltage of cell drawn has to be predicted.

Concept Introduction:

Refer to part (a).

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Answer 4

Question

Chapter 14, Problem 14.22P

(a)

Interpretation Introduction

Interpretation:

Half reactions for silver-silver chloride along with calomel electrodes used as reference have to be written.

Concept Introduction:

Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode. For a general representation the oxidation half cell is drawn in left separated by double vertical parallel lines while reduction half cell comes at right.

Expression to compute Ecell is written as follows:

Ecell°=Eright−Eleft

(b)

Interpretation Introduction

Interpretation:

Voltage of cell drawn has to be predicted.

Concept Introduction:

Refer to part (a).

Expert Solution & Answer

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Answer 5

Chapter 14, Problem 14.22P

(a)

Interpretation Introduction

Interpretation:

Half reactions for silver-silver chloride along with calomel electrodes used as reference have to be written.

Concept Introduction:

Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode. For a general representation the oxidation half cell is drawn in left separated by double vertical parallel lines while reduction half cell comes at right.

Expression to compute Ecell is written as follows:

Ecell°=Eright−Eleft

(b)

Interpretation Introduction

Interpretation:

Voltage of cell drawn has to be predicted.

Concept Introduction:

Refer to part (a).

Answer 6

Chapter 14, Problem 14.22P

(a)

Interpretation Introduction

Interpretation:

Half reactions for silver-silver chloride along with calomel electrodes used as reference have to be written.

Concept Introduction:

Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode. For a general representation the oxidation half cell is drawn in left separated by double vertical parallel lines while reduction half cell comes at right.

Expression to compute Ecell is written as follows:

Ecell°=Eright−Eleft

Answer 7

Chapter 14, Problem 14.22P

(a)

Interpretation Introduction

Interpretation:

Half reactions for silver-silver chloride along with calomel electrodes used as reference have to be written.

Concept Introduction:

Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode. For a general representation the oxidation half cell is drawn in left separated by double vertical parallel lines while reduction half cell comes at right.

Expression to compute Ecell is written as follows:

Ecell°=Eright−Eleft

Answer 8

(b)

Interpretation Introduction

Interpretation:

Voltage of cell drawn has to be predicted.

Concept Introduction:

Refer to part (a).

Answer 9

(b)

Interpretation Introduction

Interpretation:

Voltage of cell drawn has to be predicted.

Concept Introduction:

Refer to part (a).

Answer 10

Expert Solution & Answer

Answer 11

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Answer 12

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Answer 13

Ch. 14 - Prob. 14.1P Ch. 14 - Prob. 14.2P Ch. 14 - Prob. 14.3P Ch. 14 - Prob. 14.4P Ch. 14 - Prob. 14.6P Ch. 14 - Prob. 14.7P Ch. 14 - Prob. 14.8P Ch. 14 - Prob. 14.9P Ch. 14 - Prob. 14.10P Ch. 14 - Prob. 14.11P

Interpretation: Voltage of cell drawn has to be predicted. Concept Introduction: Refer to part (a).

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Chapter 14 Solutions

Interpretation:

Voltage of cell drawn has to be predicted.

Concept Introduction:

Refer to part (a).