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MnO(4)^(-) + SO(2)^(2-) + H(2)O = MnO(2)...

`MnO_(4)^(-) + SO_(2)^(2-) + H_(2)O = MnO_(2) + SO_(4)^(2-) + OH^(-)`

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To balance the redox reaction given, we will follow the half-reaction method. The reaction is: \[ \text{MnO}_4^{-} + \text{SO}_2^{2-} + \text{H}_2\text{O} \rightarrow \text{MnO}_2 + \text{SO}_4^{2-} + \text{OH}^{-} \] ### Step 1: Assign Oxidation States First, we need to assign oxidation states to identify the species being oxidized and reduced. - In \(\text{MnO}_4^{-}\), Mn has an oxidation state of +7. ...
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A solution of 0.1 M KMnO_(4) is used for the reaction S_(2)O_(3)^(2-) + 2MnO_(4)^(-) + H_(2)O to MnO_(2)+SO_(4)^(2-) + OH^(-) What volume of solution in ml will be required to react with 0.158 gm of Na_(2)S_(2)O_(3) ?

0.1 M KMnO_(4) is used for the following titration. What volume of the solution in mL will be required to react with 0.158 g of Na_(2)S_(2)O_(3) ? underset(("not balanced"))(S_(2)O_(3)^(2-))+MnO_(4)^(-)+H_(2)O to MnO_(2)(s)+SO_(4)^(2-)+OH^(-)

Knowledge Check

  • Consider the following reactions : (i) C_(2)O_(4)^(2-) rarr CO_(2) (ii) SO_(4)^(2-) rarr SO_(3)^(2-) (iii) MnO_(4)^(2-) rarr MnO_(4)^(-) (iv) Fe^(3+) rarr Fe^(2+) Choose the correct answer -

    A
    `(i) & (ii)` shows oxidation
    B
    `(iii) & (iv)` shows reduction
    C
    `(i) & (iii)` shows oxidation
    D
    `(iii) & (iv)` shows oxidation

  • For the redox reaction: MnO_(4)^(-) + C_(2)O_(4)^(2-) + H^(+) to Mn^(2+) + CO_(2) + H_(2)O The correct coefficients of the reactants for the balanced equations are: MnO_(4)^(-) -16, C_(2)O_(4)^(2-) - 5, H^(+) - 2

    A
    `MnO_(4)^(-) - 16, C_(2)O_(4)^(2-) - 5, H^(+) - 2`
    B
    `MnO_(4)^(-) - 2, C_(2)O_(4)^(2-) - 5, H^(+) - 16`
    C
    `MnO_(4)^(-) - 5, C_(2)O_(4)^(2-) - 16, H^(+) - 2`
    D
    `MnO_(4)^(-) - 2, C_(2)O_(4)^(2-) - 16, H^(+) - 5`

  • For the redox reaction: MnO_(4)^(-) + C_(2)O_(4)^(2-) + H^(+) to Mn^(2+) + CO_(2) + H_(2)O The correct coefficients of the reactants for the balanced equations are: MnO_(4)^(-) -16, C_(2)O_(4)^(2-) - 5, H^(+) - 2

    A
    `MnO_(4)^(-) - 16, C_(2)O_(4)^(2-) - 5, H^(+) - 2`
    B
    `MnO_(4)^(-) - 2, C_(2)O_(4)^(2-) - 5, H^(+) - 16`
    C
    `MnO_(4)^(-) - 5, C_(2)O_(4)^(2-) - 16, H^(+) - 2`
    D
    `MnO_(4)^(-) - 2, C_(2)O_(4)^(2-) - 16, H^(+) - 5`

    • Similar Questions

    Explore conceptually related problems

    MnO_(4)^(-) + C_(2) O_(4)^(2-) + H^(+) = CO_(2) + Mn^(2+) + H_(2) O

    Balance : KMnO_(4) + MnSO_(4) + H_(2) O to MnO_(2) + K_(2) SO_(4) + H_(2) SO_(4)

    MnO_(4)^(2-) + H^(+) = MnO_(4)^(-) + MnO_(2) + H_(2) O

    With how many of the following reagents, H_(2)O_(2) act as reducing agent. MnO_(4)^(-) H^(+) ; MnO_(4)^(-) / OH^(-) , HOCl;I_(2) / OH^(-) ;[ Fe(CN)_(6)]^(3-) / OH^(-) ;PbS;],[ Fe^(2+) / H^(+) ; Cr^(3+) / OH^(-) ]

    For the redox reaction , MnO_(4)^(-) + C_(2)O_(4)^(-2) + H^(+) to Mn^(+2) + CO_(2) + H_(2)O the correct coefficient of reactants MnO_(4)^(-), C_(2)O_(4)^(-2) H^(+) for the balanced reaction are respectively :

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