Chapter 19, Problem 105IL

Consider an electrochemical cell based on the half-reactions Ni²⁺(aq) + 2 e⁻ → Ni(s) and Cd²⁺(aq) + 2e⁻ → Cd(s).

1. (a) Diagram the cell, and label each of the components (including the anode, cathode, and salt bridge).
2. (b) Use the equations for the half-reactions to write a balanced, net ionic equation for the overall cell reaction.
3. (c) What is the polarity of each electrode?
4. (d) What is the value of E°_cell?
5. (e) In which direction do electrons flow in the external circuit?
6. (f) Assume that a salt bridge containing NaNO₃ connects the two half-cells. In which direction do the Na⁺(aq) ions move? In which direction do the NO₃⁻ (aq) ions move?
7. (g) Calculate the equilibrium constant for the reaction.
8. (h) If the concentration of Cd²⁺ is reduced to 0.010 M and [Ni²⁺] = 1.0 M, what is the value of E_cell? Is the net reaction still the reaction given in part (b)?
9. (i) If 0.050 A is drawn from the battery, how long can it last if you begin with 1.0 L of each of the solutions and each was initially 1.0 M in dissolved species? Each electrode weighs 50.0 g in the beginning.

Interpretation Introduction

Interpretation:

The half reactions are as follows.

${\text{Ni}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Ni(s) and Cd}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cd(s)}$

The cell has to be drawn and label each of the component.

Concept introduction:

Voltaic cell or Galvanic cell:

The device to produce electricity by using chemical reactions. In these divces are redox chemical reactions are occured.

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

The chemical reaction in the half cell is an oxidation reduction (redox)reactions.

For example:

Cell diagram of voltaic or galvanic cell is as follows.

$\begin{array}{l}\text{ Salt bridge}\\ \downarrow \\ {\text{Cu(s)|Cu}}^{\text{2+}}{\text{(aq) || Ag}}^{\text{+}}\text{(aq)|Ag(s)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_ \_\_\_\_\_\_\_\_\_\_\_}\\ \downarrow \downarrow \\ \text{ Half cell Half cell}\end{array}$

Answer to Problem 105IL

Explanation of Solution

The half reactions are as follows.

${\text{Ni}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Ni(s) and Cd}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cd(s)}$

The voltaic cell are as follows:

Interpretation Introduction

Interpretation:

To determine the following.

The half reactions are as follows.

${\text{Ni}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Ni(s) and Cd}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cd(s)}$

The balance equation has to be given.

Concept introduction:

Voltaic cell or Galvanic cell:

The device to produce electricity by using chemical reactions. In these divces are redox chemical reactions are occured.

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

The chemical reaction in the half cell is an oxidation reduction (redox)reactions.

For example:

Cell diagram of voltaic or galvanic cell is as follows.

$\begin{array}{l}\text{ Salt bridge}\\ \downarrow \\ {\text{Cu(s)|Cu}}^{\text{2+}}{\text{(aq) || Ag}}^{\text{+}}\text{(aq)|Ag(s)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_ \_\_\_\_\_\_\_\_\_\_\_}\\ \downarrow \downarrow \\ \text{ Half cell Half cell}\end{array}$

Answer to Problem 105IL

Balanced reaction:${\text{Cd(s) + Ni}}^{2+}(\text{aq) }\to {\text{Cd}}^{\text{2+}}\text{(aq) + Ni(s)}$

Explanation of Solution

Let’s write the half reactions occur at anode and cathode:

$\begin{array}{l}\text{At anode:}\\ \text{Cd(s) }\to {\text{Cd}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\\ \text{At cathode:}\\ {\text{Ni}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Ni(s)}\end{array}$

By adding these two half reactions we get balanced reaction.

$\begin{array}{l}\text{Cd(s) }\to {\text{Cd}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\\ {\text{Ni}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Ni(s)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{Cd(s) + Ni}}^{2+}(\text{aq) }\to {\text{Cd}}^{\text{2+}}\text{(aq) + Ni(s)}\end{array}$

Interpretation Introduction

Interpretation:

The half reactions are as follows.

${\text{Ni}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Ni(s) and Cd}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cd(s)}$

The polarity of each electrode has to be determined.

Concept introduction:

Voltaic cell or Galvanic cell:

The device to produce electricity by using chemical reactions. In these divces are redox chemical reactions are occured.

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

The chemical reaction in the half cell is an oxidation reduction (redox)reactions.

For example:

Cell diagram of voltaic or galvanic cell is as follows.

$\begin{array}{l}\text{ Salt bridge}\\ \downarrow \\ {\text{Cu(s)|Cu}}^{\text{2+}}{\text{(aq) || Ag}}^{\text{+}}\text{(aq)|Ag(s)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_ \_\_\_\_\_\_\_\_\_\_\_}\\ \downarrow \downarrow \\ \text{ Half cell Half cell}\end{array}$

Answer to Problem 105IL

Anode is negative cathode is positive.

Explanation of Solution

In the voltaic cell has two voltaic cells. One electrode has positive charge called cathode and another electrode has negative called anode.

Interpretation Introduction

Interpretation:

The half reactions are as follows.

${\text{Ni}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Ni(s) and Cd}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cd(s)}$

The ${\text{E}}_{\text{cell}}^{\text{o}}$ has to be calculated.

Concept introduction:

Voltaic cell or Galvanic cell:

The device to produce electricity by using chemical reactions. In these divces are redox chemical reactions are occured.

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

The chemical reaction in the half cell is an oxidation reduction (redox)reactions.

For example:

Cell diagram of voltaic or galvanic cell is as follows.

$\begin{array}{l}\text{ Salt bridge}\\ \downarrow \\ {\text{Cu(s)|Cu}}^{\text{2+}}{\text{(aq) || Ag}}^{\text{+}}\text{(aq)|Ag(s)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_ \_\_\_\_\_\_\_\_\_\_\_}\\ \downarrow \downarrow \\ \text{ Half cell Half cell}\end{array}$

Answer to Problem 105IL

${\text{E}}_{\text{cell}}^{\text{o}}$ of the reaction is $\text{0}\text{.15 V}$

Explanation of Solution

The reactions occur at anode and cathode is as follows.

$\begin{array}{l}\text{At anode:}\\ \text{Cd(s) }\to {\text{Cd}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= -0}\text{.40 V}\\ \text{At cathode:}\\ {\text{Ni}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to {\text{Ni(s) ; E}}^{\text{0}}\text{= -0}\text{.25 V}\end{array}$

Let’s calculate the ${\text{E}}_{\text{cell}}^{\text{o}}$ of the reaction

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{=E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}\\ \text{= -0}\text{.25 V-(-0}\text{.40) V}\\ \text{= 0}\text{.15 V}\end{array}$

Interpretation Introduction

Interpretation:

The half reactions are as follows.

${\text{Ni}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Ni(s) and Cd}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cd(s)}$

The direction in which electrons flow in the external circuit has to be given.

Concept introduction:

Voltaic cell or Galvanic cell:

The device to produce electricity by using chemical reactions. In these divces are redox chemical reactions are occured.

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

The chemical reaction in the half cell is an oxidation reduction (redox)reactions.

For example:

Cell diagram of voltaic or galvanic cell is as follows.

$\begin{array}{l}\text{ Salt bridge}\\ \downarrow \\ {\text{Cu(s)|Cu}}^{\text{2+}}{\text{(aq) || Ag}}^{\text{+}}\text{(aq)|Ag(s)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_ \_\_\_\_\_\_\_\_\_\_\_}\\ \downarrow \downarrow \\ \text{ Half cell Half cell}\end{array}$

Answer to Problem 105IL

Electrons are flow from anode to cathode.

Explanation of Solution

In the voltaic cell electrons are move anode to cathode.

Interpretation Introduction

Interpretation:

To determine the following.

The half reactions are as follows.

${\text{Ni}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Ni(s) and Cd}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cd(s)}$

Assume that a salt bride containing ${\text{NaNO}}_{\text{3}}$ connect the two half cells. It has to be identified in which direction do ${\text{Na}}^{\text{+}}$(aq) ions and ${\text{NO}}_{\text{3}}^{\text{-}}\text{ (aq)}$ move.

Concept introduction:

Voltaic cell or Galvanic cell:

The device to produce electricity by using chemical reactions. In these divces are redox chemical reactions are occured.

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

The chemical reaction in the half cell is an oxidation reduction (redox)reactions.

For example:

Cell diagram of voltaic or galvanic cell is as follows.

$\begin{array}{l}\text{ Salt bridge}\\ \downarrow \\ {\text{Cu(s)|Cu}}^{\text{2+}}{\text{(aq) || Ag}}^{\text{+}}\text{(aq)|Ag(s)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_ \_\_\_\_\_\_\_\_\_\_\_}\\ \downarrow \downarrow \\ \text{ Half cell Half cell}\end{array}$

Answer to Problem 105IL

${\text{Na}}^{\text{+}}$ ions moves from anode to cathode

${\text{NO}}_{\text{3}}^{\text{-}}$ ions move from cathode to anode

Explanation of Solution

Salt bridge contains ${\text{NaNO}}_{\text{3}}$ solution and it is dissociates into ${\text{Na}}^{\text{+}}{\text{ and NO}}_{\text{3}}^{\text{-}}$ ions.

${\text{Na}}^{\text{+}}$ ions moves from anode to cathode

${\text{NO}}_{\text{3}}^{\text{-}}$ ions move from cathode to anode

Interpretation Introduction

Interpretation:

The half reactions are as follows.

${\text{Ni}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Ni(s) and Cd}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cd(s)}$

The equilibrium constant has to be determined.

Concept introduction:

Voltaic cell or Galvanic cell:

The device to produce electricity by using chemical reactions. In these divces are redox chemical reactions are occured.

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

The chemical reaction in the half cell is an oxidation reduction (redox)reactions.

For example:

Cell diagram of voltaic or galvanic cell is as follows.

$\begin{array}{l}\text{ Salt bridge}\\ \downarrow \\ {\text{Cu(s)|Cu}}^{\text{2+}}{\text{(aq) || Ag}}^{\text{+}}\text{(aq)|Ag(s)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_ \_\_\_\_\_\_\_\_\_\_\_}\\ \downarrow \downarrow \\ \text{ Half cell Half cell}\end{array}$

Answer to Problem 105IL

The equilibrium constant of the reaction is $\text{1}{\text{.1×10}}^{\text{5}}$.

Explanation of Solution

$\begin{array}{c}\text{lnK=}\frac{{\text{nE}}^{\text{0}}}{\text{0}\text{.0257}}\\ \text{=}\frac{\text{(2)(0}\text{.15 V)}}{\text{0}\text{.0257}}\\ \text{=11}\text{.673}\\ \text{K=1}\text{.1 }\times {\text{10}}^{\text{5}}\end{array}$

Interpretation Introduction

Interpretation:

The half reactions are as follows.

${\text{Ni}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Ni(s) and Cd}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cd(s)}$

If the concentration of ${\text{Cd}}^{\text{2+}}\text{(aq)}$ is reduced to $\text{0}\text{.010 M}$ and ${\text{[Ni}}^{\text{2+}}\text{]=1}\text{.0 M}$. The ${\text{E}}_{\text{cell}}^{\text{o}}$ of the reaction has to be calculated.

Concept introduction:

Voltaic cell or Galvanic cell:

The device to produce electricity by using chemical reactions. In these divces are redox chemical reactions are occured.

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

The chemical reaction in the half cell is an oxidation reduction (redox)reactions.

For example:

Cell diagram of voltaic or galvanic cell is as follows.

$\begin{array}{l}\text{ Salt bridge}\\ \downarrow \\ {\text{Cu(s)|Cu}}^{\text{2+}}{\text{(aq) || Ag}}^{\text{+}}\text{(aq)|Ag(s)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_ \_\_\_\_\_\_\_\_\_\_\_}\\ \downarrow \downarrow \\ \text{ Half cell Half cell}\end{array}$

Answer to Problem 105IL

${\text{E}}_{\text{cell}}$ of the reaction is $\text{0}\text{.21 V}$.

Explanation of Solution

$\begin{array}{c}{\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{o}}\text{-}\frac{\text{0}\text{.0257}}{\text{2}}\text{lnQ}\\ {\text{=E}}_{\text{cell}}^{\text{o}}\text{-}\frac{\text{0}\text{.0257}}{\text{2}}\text{ln}\frac{{\text{[Cd}}^{\text{2+}}\text{]}}{{\text{[Ni}}^{\text{2+}}\text{]}}\\ \text{=0}\text{.15 V-}\frac{\text{0}\text{.0257}}{\text{2}}\text{ln}\frac{\text{(0}\text{.010 M)}}{\text{(1}\text{.0 M)}}\\ \text{=0}\text{.21 V}\end{array}$

Net reaction will be still given in part (b)

Interpretation Introduction

Interpretation:

The half reactions are as follows.

${\text{Ni}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Ni(s) and Cd}}^{\text{2+}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cd(s)}$

The time the battery will last if $\text{0}\text{.050 M}$ is drawn from the battery has to be determined.

Concept introduction:

Voltaic cell or Galvanic cell:

The device to produce electricity by using chemical reactions. In these divces are redox chemical reactions are occured.

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

The chemical reaction in the half cell is an oxidation reduction (redox)reactions.

For example:

Cell diagram of voltaic or galvanic cell is as follows.

$\begin{array}{l}\text{ Salt bridge}\\ \downarrow \\ {\text{Cu(s)|Cu}}^{\text{2+}}{\text{(aq) || Ag}}^{\text{+}}\text{(aq)|Ag(s)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_ \_\_\_\_\_\_\_\_\_\_\_}\\ \downarrow \downarrow \\ \text{ Half cell Half cell}\end{array}$

Answer to Problem 105IL

The time required for the electrolysis is $\text{4800 h}$

Explanation of Solution

$\begin{array}{c}\text{Moles of electrons required:}\\ {\text{Mole e}}^{\text{-}}\text{= (50 g Cd)}\left(\frac{\text{1 mol}}{\text{112}{\text{.4 mol e}}^{\text{-}}}\right)\left(\frac{{\text{2 mol e}}^{\text{-}}}{\text{1 mol Cd}}\right)\\ \text{=0}{\text{.88896 mol e}}^{\text{-}}\end{array}$

Let’s calculate the charge of the cell:

$\begin{array}{c}\text{Charge (C) = (0}{\text{.8896 mol e}}^{\text{-}}\text{)}\left(\frac{\text{96500 C}}{{\text{1 mol e}}^{\text{-}}}\right)\\ \text{= 8}{\text{.58 u ×10}}^{\text{4}}\text{ C}\end{array}$

Therefore, the time can be calculated as follows.

$\begin{array}{c}\text{Charge = Current (A) × time (s)}\\ \text{8}{\text{.584×10}}^{\text{4}}\text{ C= 0}\text{.05 A ×time (s)}\\ \text{time (s)=1}{\text{.7×10}}^{\text{6}}\text{ S}\\ \text{=4800 h}\end{array}$