In an experiment,$50 \ mL$ of $0.1 \ M$ solution of a metal salt reacted with $25 \ mL$ of $0.1 \ M$ solution of sodium sulphite. The half-equation for the oxidation of sulphite ion is $SO_3^{2-}{(aq)} + H_2O_{(l)} \to SO_4^{2-}{(aq)} + 2H^{+}{(aq)} + 2e^-$. If the oxidation number of the metal in the salt was $3$,what would be the new oxidation number of the metal?
- A$0$
- B$1$
- C$2$
- D$4$
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Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
$(a)$ $P_{4(s)} + OH_{(aq)}^{-} \to PH_{3(g)} + HPO_{2(aq)}^{-}$
$(b)$ $N_{2}H_{4(l)} + ClO_{3(aq)}^{-} \to NO_{(g)} + Cl_{(g)}^{-}$
$(c)$ $Cl_{2}O_{7(g)} + H_{2}O_{2(aq)} \to ClO_{2(aq)}^{-} + O_{2(g)} + H^{+}$
$(a)$ $P_{4(s)} + OH_{(aq)}^{-} \to PH_{3(g)} + HPO_{2(aq)}^{-}$
$(b)$ $N_{2}H_{4(l)} + ClO_{3(aq)}^{-} \to NO_{(g)} + Cl_{(g)}^{-}$
$(c)$ $Cl_{2}O_{7(g)} + H_{2}O_{2(aq)} \to ClO_{2(aq)}^{-} + O_{2(g)} + H^{+}$
Difficult
View Solution$1$ mole of $PbS$ is oxidised by "$X$" moles of $O_3$ to get "$Y$" moles of $O_2$. $X + Y = $ . . . . . . .
DifficultJEE MAIN 2024
View SolutionConsider the following reaction:
$xMnO_4^{-} + yC_2O_4^{2-} + zH^{+} \rightarrow xMn^{2+} + 2yCO_2 + \frac{z}{2}H_2O$
The values of $x, y$ and $z$ in the reaction are,respectively:
$xMnO_4^{-} + yC_2O_4^{2-} + zH^{+} \rightarrow xMn^{2+} + 2yCO_2 + \frac{z}{2}H_2O$
The values of $x, y$ and $z$ in the reaction are,respectively:
How many moles of hydrazine $(N_{2}H_{4})$ can be oxidized into $N_{2}$ using $\frac{2}{3}$ mole of bromate ions $(BrO_{3}^{-} \to Br^{-})$ :-
Difficult
View SolutionThe acidic solution of $S_2O_3^{2-}$ is converted to ....... in the presence of $I_2$.
Medium
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