Physics, 17.10.2019 23:30 meiyrarodriguez

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

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Answer from: shahedalahmad2017

The change in the cell voltage when the ion concentrations in the cathode half-cell are increased by a factor of 10 is 0.05 v

Further explanation

The standard reduction potentials for iron and silver are as follows:

$E^o_{(Fe^{2+}/Fe)} = -0.44 V\\E^o_{(Ag^{+}/Ag)} = 0.8 V\\$

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

The given cell reactions are:

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

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

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

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

The overall cell reaction will be,

$2Ag^{+}+Fe=Fe^{2+}+2Ag$

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

Now we have to calculate the cell potential.

Using Nernst equation:

$E_{cell}=E^0_{cell} - \frac{0.0592}{n} log \frac{[Fe^{2+}]}{[Ag^+]^2}$

where,

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

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

Now put all the given values then we get:

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

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

Hence, 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 [Ag^+] will be 10 M.

Using Nernst equation:

$E_{cell}=E^0_{cell} - \frac{0.0592}{n} log \frac{[Fe^{2+}]}{[Ag^+]^2}$

$E_{cell} = 1.24 - \frac{0.0592}{2} log \frac{1M}{(10M)^2} \\E_{cell} =1.29 V$

Then the change in cell voltage will be:

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

Thus, the change in the cell voltage when the ion concentrations in the cathode half-cell are increased by a factor of 10 is 0.05 v

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

Grade: 9

Subject: chemistry

Chapter: the change in the cell voltage

Keywords: the cathode, the cell voltage, half-cell, the change, the ion concentrations

What is the change in the cell voltage when the ion concentrations in the cathode half-cell are incr

Answer from: Quest

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006 kilo calorie

then ,802.3 kilo joule will be = 802.3 $\times$ 0.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

Answer from: Quest

im srry but ur wrong the answer is d a smooth rock, some red dirt, and an empty bird's nest.

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