Question #7cb5e

Oct 31, 2016

A

Explanation:

The cell reaction when it is working is:

$\textsf{Z n + C {u}^{2 +} r i g h t \le f t h a r p \infty n s Z {n}^{2 +} + C u}$

At $\textsf{{25}^{\circ} C}$ a working form of the Nernst Equation is:

$\textsf{{E}_{c e l l} = {E}^{\circ} - \frac{0.0592}{n} \log Q}$

$\textsf{Q}$ is the reaction quotient and is given by:

$\textsf{Q = \frac{\left[Z {n}^{2 +}\right]}{\left[C {u}^{2 +}\right]}}$

$\textsf{n}$ is the number of moles of electrons transferred which, in this case, is 2.

As current is drawn from the cell as it is working, the potential difference between the two 1/2 cells gradually falls and eventually becomes zero.

The cell is now flat and $\textsf{{E}_{c e l l} = 0}$.

Putting this in to the Nernst Equation we get:

$\textsf{0 = 1.1 - \frac{0.0592}{2} \log Q}$

$\therefore$$\textsf{\log Q = \frac{1.1 \times 2}{0.0592} = 37.16}$

This gives $\textsf{A}$ to be the correct response.

The system has now reached equilibrium at which point $\textsf{K = Q}$.

This means that $\textsf{K \cong {10}^{37}}$ which is a fantastically high number and effectively tells us that the reaction has gone to completion as you would expect from our knowledge of displacement reactions.