MCQ
A solution of sodium acetate in water will
  • Turn red litmus blue
  • B
    Turn blue litmus red
  • C
    Not effect litmus
  • D
    Decolourises litmus

Answer

Correct option: A.
Turn red litmus blue
(a) Because of $C{H_3}COONa$ is a salt of weak acid and strong base.

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$\begin{array}{*{20}{c}}
  {{{(C{H_3})}_2}CH - C{H_2}CH = CH - CH - CHC{H_3}} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2}{H_5}} 
\end{array}$

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential ( $\mathrm{E}^{\circ}$ ) 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 $E^{\circ}$ ($V$ with respect to normal hydrogen electrode) values. 

$\mathrm{I}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}\ \ \   \mathrm{E}^{\circ}=0.54 $

$\mathrm{Cl}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \ \ \  \mathrm{E}^{\circ}=1.36 $

$\mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} \ \ \  \mathrm{E}^{\circ}=1.50 $

$\mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} \ \ \  \mathrm{E}^{\circ}=0.77 $

$\mathrm{O}_2+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O} \ \ \  \mathrm{E}^{\circ}=1.23$

$1.$ Among the following, identify the correct statement.

$(A)$ Chloride ion is oxidized by $\mathrm{O}_2$

$(B)$ $\mathrm{Fe}^{2+}$ is oxidized by iodine

$(C)$ Iodide ion is oxidized by chlorine

$(D)$ $\mathrm{Mn}^{2+}$ is oxidized by chlorine

$2.$ While $\mathrm{Fe}^{3+}$ is stable, $\mathrm{Mn}^{3+}$ is not stable in acid solution because

$(A)$ $\mathrm{O}_2$ oxidises $\mathrm{Mn}^{2+}$ to $\mathrm{Mn}^{3+}$

$(B)$ $\mathrm{O}_2$ oxidises both $\mathrm{Mn}^{2+}$ and $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$

$(C)$ $\mathrm{Fe}^{3+}$ oxidizes $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$

$(D)$ $\mathrm{Mn}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$

$3.$ 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

$(A)$ $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$ $(B)$ $\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$

$(C)$ $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$ $(D)$ $\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$

Give the answer question $1,2$ and $3.$

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