Question

when the following equation is balanced properly under basic conditions, what are the coefficients of the species shown?

$\square\mathrm{cl}^{-}+\square\mathrm{i}_{2}\to\square\mathrm{i}^{-}+\square\mathrm{cl}_{2}$

water appears in the balanced equation as a $\square$ (reactant, product, neither) with a coefficient of $\square$. (enter 0 for neither.)

which element is oxidized? $\square$

Explanation:

Step1: Assign oxidation states

• For \(Cl^-\), oxidation state of \(Cl\) is \(- 1\).
• For \(I_2\), oxidation state of \(I\) is \(0\).
• For \(I^-\), oxidation state of \(I\) is \(-1\).
• For \(Cl_2\), oxidation state of \(Cl\) is \(0\).

Step2: Identify oxidation and reduction half - reactions

• Oxidation half - reaction: \(Cl^-\to Cl_2\). The change in oxidation state of \(Cl\) is from \(-1\) to \(0\).
• Reduction half - reaction: \(I_2\to I^-\). The change in oxidation state of \(I\) is from \(0\) to \(-1\).

Step3: Balance the oxidation half - reaction

• \(2Cl^-\to Cl_2 + 2e^-\) (balance atoms and charge)

Step4: Balance the reduction half - reaction

• \(I_2+2e^-\to 2I^-\) (balance atoms and charge)

Step5: Combine the half - reactions

• \(2Cl^-+I_2\to 2I^-+Cl_2\)

Answer:

• Coefficients: \(2\) (for \(Cl^-\)), \(1\) (for \(I_2\)), \(2\) (for \(I^-\)), \(1\) (for \(Cl_2\))
• Water appears as neither with a coefficient of \(0\)
• The element oxidized is \(Cl\)