Question

Electrode potential for Mg electrode varies according to the equation

$E_{Mg^{2 +}|Mg} = Eº_{Mg^{2 +}|Mg} - \frac{0.059}{2}\log\frac{1}{\lbrack Mg^{2 +}\rbrack}$

The graph of $E_{mg^{2 +}|Mg}vs\log\lbrack Mg^{2 +}\rbrack$ is

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

Answer 2

$E = Eº + \frac{0.059}{2}\log\lbrack Mg^{2 +}\rbrack$

Hence, plot of E vs $\log\lbrack Mg^{2 +}\rbrack$ will be linear with positive slope and intercept = $Eº$