(B) To find the value of $x$,we must balance the charge on both sides of the equation.In the reactant side,the total charge is: $(-1) + 6(+1) + x(-1)

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(B) To find the value of $x$,we must balance the charge on both sides of the equation.In the reactant side,the total charge is: $(-1) + 6(+1) + x(-1) = 5 - x$.In the product side,the total charge is: $(+3) + 3(0) = +3$.Equating the charges: $5 - x = 3$.Solving for $x$: $x = 5 - 3 = 2$.Thus,the balanced equation is $BiO_{3}^{-} + 6H^{+} + 2e^{-} \longrightarrow Bi^{3+} + 3H_{2}O$.

Class 11 Chemistry Redox Reactions Redox reaction and Method for balancing Redox reaction

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In the ionic equation $BiO_{3}^{-} + 6H^{+} + x e^{-} \longrightarrow Bi^{3+} + 3H_{2}O$,the value of $x$ is:

KCET 2007 All KCET

A
$6$
B
$2$
C
$4$
D
$3$

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Redox reaction and Method for balancing Redox reaction

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Assign $A$,$B$,$C$,$D$ from the given type of reaction.
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In the ionic equation $BiO_{3}^{-} + 6H^{+} + x e^{-} \longrightarrow Bi^{3+} + 3H_{2}O$,the value of $x$ is:

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$KMnO_4$ in basic medium is reduced to

Number of moles of $H^{+}$ ions required by $1 \ mole$ of $MnO_4^{-}$ to oxidise oxalate ion to $CO_2$ is . . . . . .

$C +$ Conc. $H_2SO_4 \xrightarrow{\Delta} X + Y + H_2O$. $X$ and $Y$ in the above reaction are:

$6 \times 10^{-3}$ mole $K_2Cr_2O_7$ reacts completely with $9 \times 10^{-3}$ mole $X^{n+}$ to give $XO_3^-$ and $Cr^{3+}$. The value of $n$ is

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Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$2NaI + MnO_2 + 2H_2SO_4 \to MnSO_4 + Na_2SO_4 + I_2 + 2H_2O$