Question

For the concn cell

$X_{(s)}|X^{+n}_{(aq)}(\frac{M}{20})||X^{+n}_{(aq)}(\frac{M}{2})|X_{(s)}$

$E_{cell} = 0.029V$ at $298K$ then Calculate 'n' ?

Answer 1

2

Answer 2

The given cell is a concentration cell:

$X_{(s)}|X^{+n}_{(aq)}(\frac{M}{20})||X^{+n}_{(aq)}(\frac{M}{2})|X_{(s)}$

In a concentration cell, the same electrode material is used, and the electrolyte is the same substance but at different concentrations. The standard cell potential ($E^\circ_{cell}$) for a concentration cell is zero.

The half-reactions are:

Anode (oxidation): $X_{(s)} \rightarrow X^{+n}_{(aq)}(C_1) + ne^-$

Cathode (reduction): $X^{+n}_{(aq)}(C_2) + ne^- \rightarrow X_{(s)}$

Where $C_1 = \frac{M}{20}$ (concentration in the anode compartment) and $C_2 = \frac{M}{2}$ (concentration in the cathode compartment).

For a spontaneous reaction in a concentration cell, the net reaction tends to equalize the concentrations. Thus, the higher concentration will be reduced at the cathode, and the lower concentration will be produced at the anode.

The overall cell reaction is:

$X^{+n}_{(aq)}(C_2) \rightarrow X^{+n}_{(aq)}(C_1)$

The Nernst equation for the cell at 298 K is:

$E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q$

Since $E^\circ_{cell} = 0$ for a concentration cell:

$E_{cell} = - \frac{0.0592}{n} \log Q$

The reaction quotient $Q$ for the overall reaction $X^{+n}_{(aq)}(C_2) \rightarrow X^{+n}_{(aq)}(C_1)$ is:

$Q = \frac{[X^{+n}]_{\text{anode}}}{[X^{+n}]_{\text{cathode}}} = \frac{C_1}{C_2}$

Substitute the given concentrations:

$C_1 = \frac{M}{20}$

$C_2 = \frac{M}{2}$

$Q = \frac{M/20}{M/2} = \frac{2}{20} = \frac{1}{10}$

Now substitute the value of $Q$ into the Nernst equation:

$E_{cell} = - \frac{0.0592}{n} \log \left(\frac{1}{10}\right)$

We know that $\log \left(\frac{1}{10}\right) = \log (10^{-1}) = -1$.

So, the equation becomes:

$E_{cell} = - \frac{0.0592}{n} (-1)$

$E_{cell} = \frac{0.0592}{n}$

We are given $E_{cell} = 0.029$ V.

Substitute this value into the equation:

$0.029 = \frac{0.0592}{n}$

Now, solve for 'n':

$n = \frac{0.0592}{0.029}$

$n \approx 2.041$

Since 'n' represents the number of electrons transferred in the half-reaction (or the charge/valency of the ion), it must be an integer. The calculated value $2.041$ is very close to 2. It is common in such problems that the given values or constants are slightly rounded, leading to a result that is approximately an integer.

Therefore, the value of 'n' is 2.