Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H$$_2$$ gas at the cathode is (1 Faraday = 96500 C mol$$^{-1}$$].

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential $$\left(\mathrm{E}^{\circ}\right)$$ of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $$\mathrm{E}^{\circ}$$ (V with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations.

$$\begin{array}{ll} \mathrm{I}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-} & \mathrm{E}^{\mathrm{o}}=0.54 \\ \mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} & \mathrm{E}^{\mathrm{o}}=1.36 \\ \mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} & \mathrm{E}^{\mathrm{o}}=1.50 \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} & \mathrm{E}^{\mathrm{o}}=0.77 \\ \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} & \mathrm{E}^{\mathrm{o}}=1.23 \end{array}$$

Among the following, identify the correct statement.

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential $$\left(\mathrm{E}^{\circ}\right)$$ of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $$\mathrm{E}^{\circ}$$ (V with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations.

$$\begin{array}{ll} \mathrm{I}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-} & \mathrm{E}^{\mathrm{o}}=0.54 \\ \mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} & \mathrm{E}^{\mathrm{o}}=1.36 \\ \mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} & \mathrm{E}^{\mathrm{o}}=1.50 \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} & \mathrm{E}^{\mathrm{o}}=0.77 \\ \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} & \mathrm{E}^{\mathrm{o}}=1.23 \end{array}$$

While $$\mathrm{Fe}^{3+}$$ is stable, $$\mathrm{Mn}^{3+}$$ is not stable in acid solution because

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential $$\left(\mathrm{E}^{\circ}\right)$$ of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $$\mathrm{E}^{\circ}$$ (V with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations.

$$\begin{array}{ll} \mathrm{I}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-} & \mathrm{E}^{\mathrm{o}}=0.54 \\ \mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} & \mathrm{E}^{\mathrm{o}}=1.36 \\ \mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} & \mathrm{E}^{\mathrm{o}}=1.50 \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} & \mathrm{E}^{\mathrm{o}}=0.77 \\ \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} & \mathrm{E}^{\mathrm{o}}=1.23 \end{array}$$

Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and $$\mathrm{H}_{2} \mathrm{SO}_{4}$$ in presence of air gives a Prussian blue precipitate. The blue colour is due to the formation of