Question

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

Answer 1

When the ion concentrations in the cathode half-cell are increased by a factor of 10, the change in the cell voltage is ln10 times greater. This comes from the Nernst equation. Use variation to solve for the cell voltage in terms of the initial cell voltage before increasing the ion concentrations by 10.

Answer 2

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 more

1. Learn more about the cathode brainly.com/question/11246955
2. Learn more about the cell voltage brainly.com/question/3139536
3. Learn more       about half-cell brainly.com/question/12727552

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