# For reaction $m^{x+}(aq)+xe^-\rightleftharpoons m(s)$ the nature of graph between $E_{m^{x+}/m}$ vs $\log[m^{x+}]$ is

$\begin{array}{1 1}(a)\;\text{Straight line whose slope}\;\tan\theta=\large\frac{0.051}{x}\\(b)\;\text{Parabola with vertex at origin}\\(c)\;\text{Straight line parallel to x-axis}\\(d)\;\text{none of above}\end{array}$

By Nernst equation
$E_{m^{x+}/m}=E^o_{m^{x+}/m}-\large\frac{.059}{x}$$\log \large\frac{1}{[M^{x+}]} E_{m^{x+}/m}=E^o_{m^{x+}/m}+\large\frac{0.059}{x}$$\log [M^{x+}]$
So it is straight line with slope $\large\frac{0.059}{x}$
Hence (a) is the correct answer.