Chapter 20, Problem 81A

a.

Interpretation Introduction

Interpretation:

The complete cell reaction for the given cell is to be written.

Concept introduction:

The half reactions when balanced and added make the full cell reaction.

Answer to Problem 81A

The equation for the cell reaction:

${\mathrm{Hg}}^{2+}+Sn\to S{n}^{2+}+Hg$

Explanation of Solution

Since Sn has less reduction potential than Hg, it will get oxidized resulting in this reaction.

Oxidation half reaction:

$\mathrm{Sn}\to S{n}^{2+}+2e^{-}$

Reduction half reaction:

${\mathrm{Hg}}^{2+}+2e^{-}\to Hg$

Thus, overall reaction is:

${\mathrm{Hg}}^{2+}+Sn\to S{n}^{2+}+Hg$

b.

Interpretation Introduction

Interpretation:

The substances oxidized and reduced, and the oxidizing an reducing agents are to be identified and mentioned.

Concept introduction:

The substance getting oxidized becomes reducing agent and vice versa.

Answer to Problem 81A

The oxidizing agent is mercury and the reducing agent is tin.

The substance getting oxidized is tin and the substance getting reduced is mercury.

Explanation of Solution

Since Sn has less reduction potential than Hg, it will get oxidized resulting in this reaction.

Oxidation half reaction:

$\mathrm{Sn}\to S{n}^{2+}+2e^{-}$

Reduction half reaction:

${\mathrm{Hg}}^{2+}+2e^{-}\to Hg$

$\mathrm{Sn}$ is oxidized. So, it is reducing agent.

${\mathrm{Hg}}^{2+}$ is reduced. So, it is oxidizing agent.

The deciding factor in this situation is the reduction potentials of both half cell reactions.

c.

Interpretation Introduction

Interpretation:

The reactions occurring at the anode and the cathode are to be written in balanced equation.

Concept introduction:

Oxidation occurs at the anode and reduction occurs at the cathode.

Answer to Problem 81A

The equation for anode reaction:

$\mathrm{Sn}\to {\mathrm{Sn}}^{2+}+2e^{-}$

The equation for cathode reaction:

${\mathrm{Hg}}^{2+}+2e^{-}\to \mathrm{Hg}$

Explanation of Solution

Tin is getting oxidized and oxidation occurs at the anode and mercury gets reduced and reduction occurs at the cathode.

Therefore,

Oxidation half reaction:

(Anode)

$\mathrm{Sn}\to S{n}^{2+}+2e^{-}$

Reduction half reaction:

(Cathode) ${\mathrm{Hg}}^{2+}+2e^{-}\to Hg$

d.

Interpretation Introduction

Interpretation:

The overall potential difference of the cell is to be calculated and found out.

Concept introduction : The cell potential is the sum of the potentials of its oxidation and reduction halves.

Answer to Problem 81A

The potential of the cell is 0.99V.

Explanation of Solution

$\begin{array}{l}E{°}_{oxidation}=0.14V\\ E{°}_{reduction}=0.85V\\ E{°}_{cell}=0.14+0.85=0.99V\end{array}$

e.

Interpretation Introduction

Interpretation:

The direction of movement of sulphate ions in the salt bridge is to be identified and mentioned.

Concept introduction:

In a salt bridge the positive ions move towards the side with more electrons and vice versa.

Answer to Problem 81A

The sulphate ions will move towards the mercury half cell.

Explanation of Solution

Since mercury is getting reduced there is lack of electrons in mercury half cell so sulphate ions move from the salt bridge to balance the charge.