Ncert Exercises & Solutions

The electrode potential for Mg electrode varies according to the equation

\(E_{Mg^{2+}/Mg}\ = \ E_{Mg^{2+}/Mg}^{o} \ - \ \frac{0.059}{2}log\frac{1}{[Mg^{2+}]}\)

The graph of E_{Mg²⁺ / Mg} vs log [Mg²⁺] among the following is:

1.		2.
3.		4.

HINT: Compare equation with y=c + (-mx)

Explanation:

Electrode potential for Mg electrode varies according to the equation -

\begin{array}{l} E_{{Mg}^{2 +} / Mg} = {E^{°}}_{{Mg}^{2 +} / Mg} - \frac{0.059}{2} \log \frac{1}{[{Mg}^{2 +}]} \\ E_{M g^{2 +} / M g} = {E^{°}}_{M g^{2 +} / M g} + \frac{0.059}{2} \log [{Mg}^{2 +}] \\ E_{{Mg}^{2 +} / Mg} = \frac{0.059}{2} \log [{Mg}^{2 +}] + {E^{°}}_{{Mg}^{2 +} / M g} \end{array}

This equation represents equation of straight line. It can be correlated as

\begin{matrix} E_{{Mg}^{2 +} / Mg} \\ ↑ \\ Y \end{matrix} \begin{matrix} = \end{matrix} \begin{matrix} \frac{0.059}{2} \\ ↑ \\ M \end{matrix} \begin{matrix} \log [{Mg}^{2 +}] \\ ↑ \\ X \end{matrix} \begin{matrix} + \end{matrix} \begin{matrix} {E^{°}}_{{Mg}^{2 +} / M g} \\ ↑ \\ + C \end{matrix}

So, intercept (C) = E_{Mg²⁺ / Mg}
Thus, equation can be diagramatically represented as

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