Question

when the following equation is balanced properly under basic conditions, what are the coefficients of the species shown? p + cl⁻ → cl₂ + ph₃ water appears in the balanced equation as a (reactant, product, neither) with a coefficient of (enter 0 for neither.) which element is reduced?

Explanation:

Step1: Write half - reactions

Oxidation half - reaction: $P
ightarrow PH_3$, phosphorus goes from 0 oxidation state to - 3.
Reduction half - reaction: $Cl^-
ightarrow Cl_2$, chlorine goes from - 1 oxidation state to 0.

Step2: Balance atoms other than O and H

For the oxidation half - reaction: $P + 3H^+
ightarrow PH_3$ (in acidic medium first, we will adjust for basic later).
For the reduction half - reaction: $2Cl^-
ightarrow Cl_2$.

Step3: Balance charge

Oxidation half - reaction: $P + 3H^++3e^-
ightarrow PH_3$.
Reduction half - reaction: $2Cl^-
ightarrow Cl_2 + 2e^-$.

Step4: Make electron numbers equal

Multiply oxidation half - reaction by 2 and reduction half - reaction by 3.
$2P + 6H^++6e^-
ightarrow 2PH_3$
$6Cl^-
ightarrow 3Cl_2+6e^-$

Step5: Combine half - reactions

$2P + 6Cl^-+6H^+
ightarrow 2PH_3 + 3Cl_2$

Step6: Adjust for basic conditions

Add 6 $OH^-$ to both sides:
$2P + 6Cl^-+6H^++6OH^-
ightarrow 2PH_3 + 3Cl_2+6OH^-$
Since $H^++OH^- = H_2O$, we get $2P+6Cl^- + 6H_2O
ightarrow 2PH_3+3Cl_2 + 6OH^-$

The element that is reduced is chlorine.

For the coefficients:

• Coefficient of $P$: 2
• Coefficient of $Cl^-$: 6
• Coefficient of $Cl_2$: 3
• Coefficient of $PH_3$: 2
• Water appears as a reactant with a coefficient of 6

Answer:

The element reduced is Cl.
Coefficient of $P$: 2
Coefficient of $Cl^-$: 6
Coefficient of $Cl_2$: 3
Coefficient of $PH_3$: 2
Water appears as a reactant with a coefficient of 6