What will be the change in the electrode potential of chromium electrode dipping into chromic sulphate solution, when the electrolyte is diluted 10 times at $25^\circ$C? [ $E^0\ (\text{Cr}^{3+}/\text{Cr})] = -0.74$V

The reduction half-reaction for the chromium electrode is:

$\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}$

The Nernst equation for the reduction potential is given by:

$E = E^0 - \dfrac{0.0591}{n} \log \dfrac{1}{[\text{Cr}^{3+}]}$

$E = E^0 + \dfrac{0.0591}{3} \log [\text{Cr}^{3+}]$

Let the initial concentration of $\text{Cr}^{3+}$ be $c$. The initial electrode potential is:

$E_1 = E^0 + \dfrac{0.0591}{3} \log c$

When the solution is diluted 10 times, the new concentration becomes $\dfrac{c}{10}$. The new electrode potential is:

$E_2 = E^0 + \dfrac{0.0591}{3} \log \left(\dfrac{c}{10}\right)$

The change in electrode potential is:

$\Delta E = E_2 - E_1 = \dfrac{0.0591}{3} \left[ \log \left(\dfrac{c}{10}\right) - \log c \right]$

$\Delta E = \dfrac{0.0591}{3} \log \left(\dfrac{1}{10}\right)$

$\Delta E = \dfrac{0.0591}{3} \times (-1) = -0.0197 \text{ V}$

$\Delta E = -19.7 \text{ mV}$

The negative sign indicates that the electrode potential decreases by $19.7 \text{ mV}$.

Answer: Decrease by 19.7 mV