Question

when the following equation is balanced properly under basic conditions, what are the coefficients of the species shown?
(square i_{2}+square no_{2}^{-}\tosquare i^{-}+square no_{3}^{-})
water appears in the balanced equation as a (square) (reactant, product, neither) with a coefficient of (square). (enter 0 for neither.)
which element is reduced? (square)

Explanation:

Step1: Assign oxidation numbers

• For \(I_2\), oxidation number of \(I\) is \(0\). In \(I^-\), oxidation number of \(I\) is \(- 1\).
• In \(NO_2^-\), oxidation number of \(N\) is \(+3\). In \(NO_3^-\), oxidation number of \(N\) is \(+5\).

Step2: Write half - reactions

• Reduction half - reaction: \(I_2

ightarrow I^-\)

• Balance \(I\) atoms: \(I_2

ightarrow 2I^-\)

• Balance charge: \(I_2 + 2e^-

ightarrow 2I^-\)

• Oxidation half - reaction: \(NO_2^-

ightarrow NO_3^-\)

• Balance \(O\) atoms (in basic medium, add \(OH^-\)): \(NO_2^-+2OH^-

ightarrow NO_3^- + H_2O\)

• Balance \(H\) atoms (already balanced in the above step for \(O\) in basic medium)
• Balance charge: \(NO_2^-+2OH^-

ightarrow NO_3^- + H_2O+2e^-\)

Step3: Combine half - reactions

• Since the number of electrons in both half - reactions (\(2e^-\)) is the same, add them:

\(I_2+NO_2^- + 2OH^-
ightarrow 2I^-+NO_3^-+H_2O\)

Answer:

• Coefficients: \(1I_2 + 1NO_2^-

ightarrow 2I^-+1NO_3^-\)

• Water appears as a product with a coefficient of \(1\)
• The element that is reduced is \(I\) (iodine)