Question

Electrode potential for Mg electrode varies according to the equation:

$E_{M{g}^{2+}/Mg}=E_{M{g}^{2+}/Mg}^0-\frac{0.059}{2}\log \frac{1}{\left[M{g}^{2+}\right]}$

The graph of $E_{M{g}^{2+}/Mg}$ vs $\log \left[M{g}^{2+}\right]$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Relation between $E_{M{g}^{2+}/Mg}$ and $\log \left[M{g}^{2+}\right]$

$E_{M{g}^{2+}/Mg}=E_{M{g}^{2+}/Mg}^0-\frac{0.059}{2}\log \frac{1}{\left[M{g}^{2+}\right]}$

$E_{M{g}^{2+}/Mg}=E_{M{g}^{2+}/Mg}^0-\frac{0.059}{2}\log \left[M{g}^{2+}\right]$

We need to compare this equation with a straight-line equation and draw a graph, where $y-$ coordinate denotes $E_{m{g}^{2+}/Mg}$ and $x-$coordinate denotes $\log \left[M{g}^{2+}\right]$.

Comparing the equation with a straight line equation:

$y=mx+c$
Where, $m=\text{slope},c=y-\text{intercept}$

We get,
$y=E_{M{g}^{2+}/Mg},x=\log \left[M{g}^{2+}\right]$
$c=E_{M{g}^{2+}/Mg},m=\frac{0.059}{2}$(a constant)

Thus, the graph between $E_{M{g}^{2+}/Mg}$ and $\log \left[M{g}^{2+}\right]$ will give a linearly increasing straight line with positive slope and positive $y$− intercept.


Hence, the correct answer is option(b)