(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 Reactions

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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:

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In the balanced reaction $xBrO_3^- + yCr^{3+} + zH_2O \to aBr_2 + bHCrO_4^- + cH^+$,the coefficients $x, y$ and $z$ are

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