1
KCET 2023
MCQ (Single Correct Answer)
+1
-0

During the electrolysis of brine, by using inert electrodes,

A
$$\mathrm{O}_2$$ liberates at anode
B
$$\mathrm{H}_2$$ liberates at anode
C
Na deposits on cathode
D
$$\mathrm{Cl}_2$$ liberates at anode
2
KCET 2023
MCQ (Single Correct Answer)
+1
-0

Consider the following 4 electrodes

$$\begin{aligned} & \mathrm{A}: \mathrm{Ag}^{+}(0.0001 \mathrm{M}) / \mathrm{Ag}(s) ; \\ & \mathrm{B}: \mathrm{Ag}^{+}(0.1 \mathrm{M}) / \mathrm{Ag}(s) ; \\ & \mathrm{C}: \mathrm{Ag}^{+}(0.01 \mathrm{M}) / \mathrm{Ag}(s) ; \\ & \mathrm{D}: \mathrm{Ag}^{+}(0.001 \mathrm{M}) / \mathrm{Ag}(s) ; E^{\circ}{ }_{\mathrm{Ag}^{+} / \mathrm{Ag}}=+0.80 \mathrm{~V} \end{aligned}$$

Then reduction potential in volts of the electrodes in the order.

A
B > C > D > A
B
C > D > A > B
C
A > D > C > B
D
A > B > C > D
3
KCET 2023
MCQ (Single Correct Answer)
+1
-0

When $$\mathrm{FeCl}_3$$ is added to excess of hot water gives a sol '$$X$$'. When $$\mathrm{FeCl}_3$$ is added to $$\mathrm{NaOH}(a q)$$ solution, gives sol '$$Y$$'

$$X$$ and $$Y$$ formed in the above processes respectively are

A
$$\mathrm{Fe}_2 \mathrm{O}_3 \cdot x \mathrm{H}_2 \mathrm{O} / \mathrm{OH}^{-}$$ and $$\mathrm{Fe}_2 \mathrm{O}_3 \cdot x \mathrm{H}_2 \mathrm{O} / \mathrm{Fe}^{3+}$$
B
$$\mathrm{Fe}_2 \mathrm{O}_3 \cdot x \mathrm{H}_2 \mathrm{O} / \mathrm{H}^{+}$$ and $$\mathrm{Fe}_2 \mathrm{O}_3 \cdot x \mathrm{H}_2 \mathrm{O} / \mathrm{Na}^{+}$$
C
$$\mathrm{Fe}_2 \mathrm{O}_3 \cdot x \mathrm{H}_2 \mathrm{O} / \mathrm{Cl}^{-}$$ and $$\mathrm{Fe}_2 \mathrm{O}_3 \cdot x \mathrm{H}_2 \mathrm{O} / \mathrm{OH}^{-}$$
D
$$\mathrm{Fe}_2 \mathrm{O}_3 \cdot x \mathrm{H}_2 \mathrm{O} / \mathrm{Fe}^{3+}$$ and $$\mathrm{Fe}_2 \mathrm{O}_3 \cdot x \mathrm{H}_2 \mathrm{O} / \mathrm{OH}^{-}$$
4
KCET 2023
MCQ (Single Correct Answer)
+1
-0

The reducing agent in the given equations

$$4 \mathrm{Ag}(s)+8 \mathrm{CN}^{-}(a q)+2 \mathrm{H}_2 \mathrm{O}(a q)+\mathrm{O}_2(g) \longrightarrow 4\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-}(a q)+4 \mathrm{OH}^{-}(a q)$$

$$2\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-}(a q)+\mathrm{Zn}(s) \longrightarrow {\left[\mathrm{Zn}(\mathrm{CN})_4\right]^{2-}(a q)+2 \mathrm{Ag}(s)}$$

A
$$\mathrm{Zn}$$
B
$$\mathrm{O}_2$$
C
$$\mathrm{H}_2 \mathrm{O}$$
D
$$\mathrm{CN}^{-}$$
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