Question

The following equilibrium reaction is given at 25°C

$Cu^{2+}(aq) + Sn^{2+}(aq) \rightleftharpoons Sn^{4+}(aq) + Cu(s)$

$E^\circ_{Cu^{2+}/Cu} = 0.34 \ V$

$E^\circ_{Sn^{4+}/Sn^{2+}} = 0.15 \ V$

If the gibbs free energy change for this reaction is X kJ/mol. Then calculate the value of $\frac{X}{6}$ (magnitude only) to nearest integer.

Answer 1

6

Answer 2

1. Identify the reduction (cathode) and oxidation (anode) half-reactions from the given overall reaction.

2. Determine the number of electrons (n) transferred in the balanced reaction.

3. Calculate the standard cell potential ($E^\circ_{cell}$) using the formula $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$, where $E^\circ_{cathode}$ and $E^\circ_{anode}$ are the standard reduction potentials of the respective half-reactions.

4. Calculate the standard Gibbs free energy change ($\Delta G^\circ$) using the formula $\Delta G^\circ = -nFE^\circ_{cell}$.

5. The calculated $\Delta G^\circ$ is X kJ/mol. Take the magnitude of X and divide it by 6.

6. Round the result to the nearest integer.

Solution:

The given equilibrium reaction is:

$Cu^{2+}(aq) + Sn^{2+}(aq) \rightleftharpoons Sn^{4+}(aq) + Cu(s)$

The standard electrode potentials are:

$E^\circ(Cu^{2+}/Cu) = 0.34 \ V$

$E^\circ(Sn^{4+}/Sn^{2+}) = 0.15 \ V$

First, let's identify the oxidation and reduction half-reactions from the given overall reaction:

1. Reduction half-reaction: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

   The standard reduction potential for this half-reaction is $E^\circ(Cu^{2+}/Cu) = 0.34 \ V$. This occurs at the cathode.

2. Oxidation half-reaction: $Sn^{2+}(aq) \rightarrow Sn^{4+}(aq) + 2e^-$

   The standard reduction potential for the $Sn^{4+}/Sn^{2+}$ couple is $E^\circ(Sn^{4+}/Sn^{2+}) = 0.15 \ V$. This means that $Sn^{4+} + 2e^- \rightarrow Sn^{2+}$ has $E^\circ = 0.15 \ V$. Since $Sn^{2+}$ is oxidized to $Sn^{4+}$ in our reaction, this occurs at the anode.

From the half-reactions, we can see that the number of electrons transferred (n) is 2.

Next, calculate the standard cell potential ($E^\circ_{cell}$):

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$

Using standard reduction potentials:

$E^\circ_{cell} = E^\circ(Cu^{2+}/Cu) - E^\circ(Sn^{4+}/Sn^{2+})$

$E^\circ_{cell} = 0.34 \ V - 0.15 \ V$

$E^\circ_{cell} = 0.19 \ V$

Now, calculate the standard Gibbs free energy change ($\Delta G^\circ$) for the reaction using the formula:

$\Delta G^\circ = -nFE^\circ_{cell}$

Where:

n = number of electrons transferred = 2

F = Faraday constant = 96485 C/mol

$E^\circ_{cell}$ = standard cell potential = 0.19 V

$\Delta G^\circ = - (2 \ mol \ e^-) \times (96485 \ C/mol \ e^-) \times (0.19 \ J/C)$

$\Delta G^\circ = - 36664.3 \ J$

Convert Joules to kilojoules:

$\Delta G^\circ = - 36664.3 \ J / 1000 \ J/kJ$

$\Delta G^\circ = - 36.6643 \ kJ/mol$

The problem states that the Gibbs free energy change for this reaction is X kJ/mol.

So, X = -36.6643 kJ/mol.

We need to calculate the value of X/6 (magnitude only) to the nearest integer.

Magnitude of X = |-36.6643| = 36.6643

Now, calculate |X|/6:

|X|/6 = 36.6643 / 6

|X|/6 = 6.110716...

Rounding this value to the nearest integer gives 6.