Question

What will be the correct balanced redox reaction for the given equation? (Use oxidation number method)
$T{l}_2{O}_3\left(s\right)+N{H}_{2}OH\left(aq\right)\to TlOH\left(s\right)+N_2\left(g\right)$

• A. $T{l}_2{O}_3\left(s\right)+2N{H}_{2}OH\left(aq\right)\to 2TlOH\left(s\right)+N_2\left(g\right)+5H_{2}O$
• B. $2T{l}_2{O}_3\left(s\right)+4N{H}_{2}OH\left(aq\right)\to 4TlOH\left(s\right)+2N_2\left(g\right)+5H_{2}O$
• C. $T{l}_2{O}_3\left(s\right)+4N{H}_{2}OH\left(aq\right)\to 2TlOH\left(s\right)+2N_2\left(g\right)+5H_{2}O$
• D. $T{l}_2{O}_3\left(s\right)+2N{H}_{2}OH\left(aq\right)\to 2TlOH\left(s\right)+N_2\left(g\right)+5H_{2}O$

Answer 1

The correct option is C $T{l}_2{O}_3\left(s\right)+4N{H}_{2}OH\left(aq\right)\to 2TlOH\left(s\right)+2N_2\left(g\right)+5H_{2}O$

$T{l}_2{O}_3\left(s\right)+N{H}_{2}OH\left(aq\right)\to TlOH\left(s\right)+N_2\left(g\right)$

Oxidation state of $Tl$ in $T{l}_2{O}_3=+3$
Oxidation state of $Tl$ in $TlOH=+1$
Oxidation state of $N$ in $N{H}_{2}OH=-1$
Oxidation state of $N$ in $N_2=0$

$N{H}_{2}OH$ is undergoing oxidation.So it acts as a reducing agent.
$T{l}_2{O}_3$ is undergoing reduction.So it acts as an oxidising agent.
$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atom}$

Finding $n_f:$
$n_f$ of $T{l}_2{O}_3=4$
$n_f$ of $N{H}_{2}OH=1$

Cross multiplying these with $n_f$ of each other.
we get,
$T{l}_2{O}_3\left(s\right)+4N{H}_{2}OH\left(aq\right)\to TlOH\left(s\right)+N_2\left(g\right)$

Balancing the elements other than oxygen and hydrogen on both sides,
$T{l}_2{O}_3\left(s\right)+4N{H}_{2}OH\left(aq\right)\to 2TlOH\left(s\right)+2N_2\left(g\right)$

Adding the $H_{2}O$ to balance the oxygen,

$T{l}_2{O}_3\left(s\right)+4N{H}_{2}OH\left(aq\right)\to 2TlOH\left(s\right)+2N_2\left(g\right)+5H_{2}O$

Hydrogen is already balanced.
Charges on both sides is also balanced.
So, it is the final balanced equation.