Question

Electrode potential for Mg electrode varies according to the equation
$E_{M{g}^{2+}/Mg}=E_{M{g}^{2+}/Mg}^O-\frac{0.059}{2}\log \frac{1}{\left[M{g}^{2+}\right]}$
The graph of $E_{M{g}^{2+}/Mg}$ versus $\log \left(M{g}^{2+}\right)$ is

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

The given reaction can be written as,
$E=E_{cell}^0+\frac{0.059}{2}\log \left[M{g}^{2+}\right]$
On comparing this with standard equation of straight line $y=mx+c$
We can make out that the plot of $E$ vs $\log$ and $\left[M{g}^{2+}\right]$ will be linear with positive slope and intercept $=E_0$