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Q.
In the ionic equation, $BiO _{3}^{-}+6 H ^{+}+x e^{-} \longrightarrow Bi ^{3+}+3 H _{2} O$, the value of $x$ is
Redox Reactions
Solution:
$BiO _{3}^{-}+6 H ^{+}+x e^{-} \longrightarrow Bi ^{3+}+3 H _{2} O$
The half cell reaction can be written as :
$Bi ^{5+} \xrightarrow{+2 e^{-}} Bi ^{3+}$
$\overset{+5} {Bi}\overset{-2}{O _{3}^{-}}+6 H ^{+}+2 e^{-} \longrightarrow Bi ^{3+}+3 H _{2} O$
Thus, the value of $x=2$ in the above ionic equation. Also, by another way
$BO _{3}^{-}+6 H ^{+}+x e^{-} \longrightarrow B ^{3+}+3 H _{2} O$
Total charge on $LHS =$ Total charge on $RHS$
$-1+6+x(-1) =+3$
$-x =+3-5$
$-x =-2 \Rightarrow x=2$