Chemistry Smartwork Access Code Fourth Edition
Chemistry Smartwork Access Code Fourth Edition
4th Edition
ISBN: 9780393521368
Author: Gilbert
Publisher: NORTON
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Chapter 19, Problem 19.36QP

(a)

Interpretation Introduction

Interpretation: The balanced equation for the cell reaction for each of the given pairs is to be stated. The identification of the half reaction at each anode and cathode is to be done.

Concept introduction: The half cell in which oxidation takes place is known as anode and the half cell at which reduction takes place is called cathode.

To determine: The balanced equation for the cell reaction for each of the given pair and the identification of the half reaction at each anode and cathode.

(a)

Expert Solution
Check Mark

Answer to Problem 19.36QP

Solution

The balanced equation for the cell reaction for each of the given pair is shown in equation (5). The half reaction at anode is shown in equation (3) and the half reaction at cathode is shown in equation (4).

Explanation of Solution

Explanation

The given pair is,

Cd2+(aq)+2eCd(s)Ag+(aq)+eAg(s)

The standard reduction potentials are,

Cd2+(aq)+2eCd(s)Eredο=0.403V (1)

Ag+(aq)+eAg(s)Eredο=0.7996V (2)

Equation (1) has more Eredο value. Therefore, reverse the equation (1) to get positive value of Ecellο .

Cd(s)Cd2+(aq)+2eEoxο=0.403V (3)

Multiply equation (2) by 2 for balancing the electrons.

2Ag+(aq)+2e2Ag(s)Eredο=0.7996V (4)

Add equation (3) and (4) to get the balanced equation.

2Ag+(aq)+Cd(s)2Ag(s)+Cd2+(aq) (5)

The half reaction at anode is shown in equation (3) and the half reaction at cathode is shown in equation (4).

(b)

Interpretation Introduction

To determine: The balanced equation for the cell reaction for each of the given pair and the identification of the half reaction at each anode and cathode.

(b)

Expert Solution
Check Mark

Answer to Problem 19.36QP

Solution

The balanced equation for the cell reaction for each of the given pair is shown in equation (10). The half reaction at anode is shown in equation (8) and the half reaction at cathode is shown in equation (7).

Explanation of Solution

Explanation

The given pair is,

AgBr(s)+eAg(s)+Br(aq)MnO2(s)+4H+(aq)+2eMn2+(aq)+2H2O(l)

The standard reduction potentials are,

AgBr(s)+eAg(s)+Br(aq)Eredο=0.095V (6)

MnO2(s)+4H+(aq)+2eMn2+(aq)+2H2O(l)Eredο=1.23V (7)

Reverse the equation (6) to get positive value of Ecellο .

Ag(s)+Br(aq)AgBr(s)+eEoxο=0.095V (8)

Multiply equation (8) by 2 for balancing the electrons.

2Ag(s)+2Br(aq)2AgBr(s)+2eEredο=0.095V (9)

Add equation (7) and (9) to get the balanced equation.

2Ag(s)+2Br(aq)+MnO2(s)+4H+(aq)2AgBr(s)+Mn2+(aq)+2H2O(l) (10)

The half reaction at anode is shown in equation (8) and the half reaction at cathode is shown in equation (7).

(c)

Interpretation Introduction

To determine: The balanced equation for the cell reaction for each of the given pair and the identification of the half reaction at each anode and cathode.

(c)

Expert Solution
Check Mark

Answer to Problem 19.36QP

Solution

The balanced equation for the cell reaction for each of the given pair is shown in equation (15). The half reaction at anode is shown in equation (13) and the half reaction at cathode is shown in equation (11).

Explanation of Solution

Explanation

The given pair is,

PtCl42(aq)+2ePt(s)+4Cl(aq)AgCl(s)+eAg(s)+Cl(aq)

The standard reduction potentials are,

PtCl42(aq)+2ePt(s)+4Cl(aq)Eredο=0.73V (11)

AgCl(s)+eAg(s)+Cl(aq)Eredο=0.2223V (12)

Reverse the equation (12) to get positive value of Ecellο .

Ag(s)+Cl(aq)AgCl(s)+eEoxο=0.2223V (13)

Multiply equation (13) by 2 for balancing the electrons.

2Ag(s)+2Cl(aq)2AgCl(s)+2eEoxο=0.2223V (14)

Add equation (11) and (14) to get the balanced equation.

2Ag(s)+2Cl(aq)+PtCl42(aq)2AgCl(s)+Pt(s)+4Cl(aq) (15)

The half reaction at anode is shown in equation (13) and the half reaction at cathode is shown in equation (11).

Conclusion

The balanced equations for the cell reaction for each of the given pairs have been stated

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Chapter 19 Solutions

Chemistry Smartwork Access Code Fourth Edition

Ch. 19 - Prob. 19.3VPCh. 19 - Prob. 19.4VPCh. 19 - Prob. 19.5VPCh. 19 - Prob. 19.6VPCh. 19 - Prob. 19.7VPCh. 19 - Prob. 19.8VPCh. 19 - Prob. 19.9VPCh. 19 - Prob. 19.10VPCh. 19 - Prob. 19.11QPCh. 19 - Prob. 19.12QPCh. 19 - Prob. 19.13QPCh. 19 - Prob. 19.14QPCh. 19 - Prob. 19.15QPCh. 19 - Prob. 19.16QPCh. 19 - Prob. 19.17QPCh. 19 - Prob. 19.18QPCh. 19 - Prob. 19.19QPCh. 19 - Prob. 19.20QPCh. 19 - Prob. 19.21QPCh. 19 - Prob. 19.22QPCh. 19 - Prob. 19.23QPCh. 19 - Prob. 19.24QPCh. 19 - Prob. 19.25QPCh. 19 - Prob. 19.26QPCh. 19 - Prob. 19.27QPCh. 19 - Prob. 19.28QPCh. 19 - Prob. 19.29QPCh. 19 - Prob. 19.30QPCh. 19 - Prob. 19.31QPCh. 19 - Prob. 19.32QPCh. 19 - Prob. 19.33QPCh. 19 - Prob. 19.34QPCh. 19 - Prob. 19.35QPCh. 19 - Prob. 19.36QPCh. 19 - Prob. 19.37QPCh. 19 - Prob. 19.38QPCh. 19 - Prob. 19.39QPCh. 19 - Prob. 19.40QPCh. 19 - Prob. 19.41QPCh. 19 - Prob. 19.42QPCh. 19 - Prob. 19.43QPCh. 19 - Prob. 19.44QPCh. 19 - Prob. 19.45QPCh. 19 - Prob. 19.46QPCh. 19 - Prob. 19.47QPCh. 19 - Prob. 19.48QPCh. 19 - Prob. 19.49QPCh. 19 - Prob. 19.50QPCh. 19 - Prob. 19.51QPCh. 19 - Prob. 19.52QPCh. 19 - Prob. 19.53QPCh. 19 - Prob. 19.54QPCh. 19 - Prob. 19.55QPCh. 19 - Prob. 19.56QPCh. 19 - Prob. 19.57QPCh. 19 - Prob. 19.58QPCh. 19 - Prob. 19.59QPCh. 19 - Prob. 19.60QPCh. 19 - Prob. 19.61QPCh. 19 - Prob. 19.62QPCh. 19 - Prob. 19.63QPCh. 19 - Prob. 19.64QPCh. 19 - Prob. 19.65QPCh. 19 - Prob. 19.66QPCh. 19 - Prob. 19.67QPCh. 19 - Prob. 19.68QPCh. 19 - Prob. 19.69QPCh. 19 - Prob. 19.70QPCh. 19 - Prob. 19.71QPCh. 19 - Prob. 19.72QPCh. 19 - Prob. 19.73QPCh. 19 - Prob. 19.74QPCh. 19 - Prob. 19.75QPCh. 19 - Prob. 19.76QPCh. 19 - Prob. 19.77QPCh. 19 - Prob. 19.78QPCh. 19 - Prob. 19.79QPCh. 19 - Prob. 19.80QPCh. 19 - Prob. 19.81QPCh. 19 - Prob. 19.82QPCh. 19 - Prob. 19.83QPCh. 19 - Prob. 19.84QPCh. 19 - Prob. 19.85QPCh. 19 - Prob. 19.86QPCh. 19 - Prob. 19.87QPCh. 19 - Prob. 19.88QPCh. 19 - Prob. 19.89APCh. 19 - Prob. 19.90APCh. 19 - Prob. 19.91APCh. 19 - Prob. 19.92AP
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