Chemistry, 30.01.2020 23:01 pstiner

A) what is the change in the cell voltage when the ion concentrations in the cathode half-cell are increased by a factor of 10? b) what is the change in the cell voltage when the ion concentrations in the anode half-cell are increased by a factor of 10?

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Answer from: deidrabenson72331

(a) The change in cell voltage is 0.05V

(b) The change in cell voltage is 0.03V

Explanation:

Redox reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases.

Further explanation:

The image taken in context is attached below.

The standard reduction potentials for iron and silver are:

$E^o_{(Fe^{2+}/Fe)}=-0.44V\\E^o_{(Ag^{+}/Ag)}=+0.80V$

In the given cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

So, silver will undergo reduction reaction will get reduced. Iron will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode): $Fe\rightarrow Fe^{2+}+2e^-$

Reduction half reaction (cathode): $Ag^{+}+e^-\rightarrow Ag$

Thus, the anode and cathode will be $E^o_{(Fe^{2+}/Fe)}$

and $E^o_{(Ag^{+}/Ag)}$ respectively.

The overall cell reaction will be,

$2Ag^{+}+Fe\rightarrow Fe^{2+}+2Ag$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{(Ag^{+}/Ag)}-E^o_{(Fe^{2+}/Fe)}$

E^o=(+0.80V)-(-0.44V)=1.24V

Now we have to calculate the cell potential.

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]}{[Ag^{+}]^2}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=1.24-\frac{0.0592}{2}\log \frac{(1M)}{(1M)^2}$

$E_{cell}=1.24V$

Thus, the emf of cell potential is 1.24 V

Part (a):

The ion concentrations in the cathode half-cell (Ag^+/Ag) are increased by a factor of 10 from 1 M to 10 M.

The emf of the cell potential will be,

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]}{[Ag^{+}]^2}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=1.24-\frac{0.0592}{2}\log \frac{(1M)}{(10M)^2}$

$E_{cell}=1.29V$

The change in cell voltage will be,

$E_{cell}=1.29V-1.24V=0.05V$

Thus, the change in cell voltage is 0.05V

Part (b):

The ion concentrations in the anode half-cell $(Fe^{2+}/Fe)$ are increased by a factor of 10 from 1 M to 10 M.

The emf of the cell potential will be,

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]}{[Ag^{+}]^2}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=1.24-\frac{0.0592}{2}\log \frac{(10M)}{(1M)^2}$

$E_{cell}=1.27V$

The change in cell voltage will be,

$E_{cell}=1.27V-1.24V=0.03V$

Thus, the change in cell voltage is 0.03V

Learn more:

Spontaneity of reaction; (answer by Kobenhavn)

Standard reduction potential; (answer by RomeliaThurston)

Keywords:

Nernst equation, standard reduction potential, spontaneity of the reaction.

A) what is the change in the cell voltage when the ion concentrations in the cathode half-cell are i

Answer from: 0Me0

The change in the cell voltage when the ion concentrations in the ANODE half-cell are increased by a factor of 10 is 0.030 V

Further explanation

What is the change in the cell voltage when the ion concentrations in the ANODE half-cell are increased by a factor of 10?

The Nernst equation

$E=E^o - \frac{RT}{nF} lnQ$

E ∘ is the cell potential at standard conditions R is the ideal gas constant T is the absolute temperature n is the number of electrons transferred per mole of reaction F is the Faraday constant Q is the reaction quotient of the reaction

The given electrochemical cell has both aqueous species Fe 2 + and Ag + at 1 M concentration

If the anode concentration is increased by a factor of 10 , the cell potential will change by the correction term:

$\Delta E = -\frac{RT}{nF} lnQ$

Now we determine the overall reaction

$Fe(s) - Fe^{2+}(aq)+2e [anode]\\2*(Ag^+(aq)+e - Ag(s)) [cathode]\\Fe(s)+2Ag^+(aq)-Fe^{2+}(aq)+2Ag(s) [overall]$

n=2 electrons were transferred

$Q=\frac{[Fe^{2+}]}{[Ag^+]^2}$

The change in cell potential is

$\Delta E = - \frac{RT}{nF} ln \frac{[Fe^{2+}]}{[Ag^+]^2}$

Note that $[Fe^{2+}]=10M$ and assume that T=298.15 K

$\Delta E = - \frac{8.314J/molK*298.15K}{2*96485C/mol} ln \frac{10M}{1M^2}\\\Delta E = -0.030 V$

Therefore the cell potential will decrease by 0.030 V

Learn moreLearn more about the cell voltage Learn more about the ion concentrations Learn more about the cathode half-cell Answer details

Grade: 9

Subject: chemistry

Chapter: the ion concentrations

Keywords: the cell voltage, the ion concentrations, the cathode half-cell, the anode, factor

Answer from: Quest

b.final

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

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