$50 \ mL$ of $0.1 \ M$ solution of a salt reacted with $25 \ mL$ of $0.1 \ M$ solution of sodium sulphite. The half-reaction for the oxidation of sulphite ion is:
$SO_{3}^{2-} (aq) + H_{2}O (l) \to SO_{4}^{2-} (aq) + 2H^{+} (aq) + 2e^{-}$
If the oxidation number of the metal in the salt was $+3$,what would be the new oxidation number of the metal?
$SO_{3}^{2-} (aq) + H_{2}O (l) \to SO_{4}^{2-} (aq) + 2H^{+} (aq) + 2e^{-}$
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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Similar Questions
Which of the following is not correct?
$1 \text{ mole of } H_2C_2O_4 \text{ is oxidised by } x \text{ mole of } MnO_4^- \text{ in strong basic medium and } 1 \text{ mole of } NaHC_2O_4 \text{ is oxidised by } y \text{ mole of } MnO_4^- \text{ in acidic medium. Ratio of } x/y \text{ is}$
Difficult
View Solution$I^{-}(aq.) + MnO_4^{-}(aq.) \xrightarrow{H^{+}} X + Mn^{2+}(aq.)$
$I^{-}(aq.) + MnO_4^{-}(aq.) \xrightarrow{\text{Neutral or weakly } OH^{-}} Y + MnO_2$
$MnO_4^{-}(aq.) + Mn^{2+}(aq.) \xrightarrow{ZnSO_4} Z + 4H^{+}$
Products $X, Y$ and $Z$ are respectively :
$I^{-}(aq.) + MnO_4^{-}(aq.) \xrightarrow{\text{Neutral or weakly } OH^{-}} Y + MnO_2$
$MnO_4^{-}(aq.) + Mn^{2+}(aq.) \xrightarrow{ZnSO_4} Z + 4H^{+}$
Products $X, Y$ and $Z$ are respectively :
Difficult
View SolutionWhat is obtained when solutions of $KI$ and $CuSO_4$ are mixed?
Medium
View SolutionConsider the reactions:
$2S_2O_3^{2-}{(aq)} + I_{2(s)} \to S_4O_6^{2-}{(aq)} + 2I^{-}{(aq)}$
$S_2O_3^{2-}{(aq)} + 2Br_{2(l)} + 5H_2O_{(l)} \to 2SO_4^{2-}{(aq)} + 4Br^{-}{(aq)} + 10H^{+}{(aq)}$
Why does the same reductant,thiosulphate,react differently with iodine and bromine?
$2S_2O_3^{2-}{(aq)} + I_{2(s)} \to S_4O_6^{2-}{(aq)} + 2I^{-}{(aq)}$
$S_2O_3^{2-}{(aq)} + 2Br_{2(l)} + 5H_2O_{(l)} \to 2SO_4^{2-}{(aq)} + 4Br^{-}{(aq)} + 10H^{+}{(aq)}$
Why does the same reductant,thiosulphate,react differently with iodine and bromine?
Medium
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