Question

balance the reaction between ph₃ and cl₂ to form p and cl⁻ in basic solution. when you have balanced the equation using the smallest integers possible, enter the coefficients of the species shown. ph₃ + cl₂ → p + cl⁻. water appears in the balanced equation as a (reactant, product, neither) with a coefficient of (enter 0 for neither). how many electrons are transferred in this reaction?

Explanation:

Step1: Write half - reactions

Oxidation: $\text{PH}_3
ightarrow\text{P}+3\text{H}^+$ (in acidic medium first, we will adjust for basic later). The oxidation state of P changes from - 3 in $\text{PH}_3$ to 0 in P, so the half - reaction for oxidation is $\text{PH}_3
ightarrow\text{P}+3\text{H}^++3e^-$.
Reduction: $\text{Cl}_2 + 2e^-
ightarrow2\text{Cl}^-$.

Step2: Balance electrons

To balance the electrons, we multiply the oxidation half - reaction by 2 and the reduction half - reaction by 3.
Oxidation: $2\text{PH}_3
ightarrow2\text{P}+6\text{H}^++6e^-$
Reduction: $3\text{Cl}_2+6e^-
ightarrow6\text{Cl}^-$

Step3: Combine half - reactions

Combining gives $2\text{PH}_3 + 3\text{Cl}_2
ightarrow2\text{P}+6\text{Cl}^-+6\text{H}^+$

Step4: Adjust for basic solution

Since it is a basic solution, we add 6 $\text{OH}^-$ to both sides to neutralize the $\text{H}^+$.
$2\text{PH}_3 + 3\text{Cl}_2+6\text{OH}^-
ightarrow2\text{P}+6\text{Cl}^-+6\text{H}_2\text{O}$
The balanced equation is $2\text{PH}_3 + 3\text{Cl}_2+6\text{OH}^-
ightarrow2\text{P}+6\text{Cl}^-+6\text{H}_2\text{O}$

Step5: Determine the number of electrons transferred

From the balanced half - reactions, the number of electrons transferred is 6.

Answer:

The coefficients of the species are: 2 for $\text{PH}_3$, 3 for $\text{Cl}_2$, 2 for $\text{P}$, 6 for $\text{Cl}^-$, 6 for $\text{OH}^-$ and 6 for $\text{H}_2\text{O}$. The number of electrons transferred is 6.