Question

Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reaction mixture has pH = 0.0, [Cr2O72-] = 1.0 M, [Cl-] = 1.0 M, [Cr3+] = 0.10 M, and parital pressure of chlorine gas of 0.010 atm.

Answer 1

Value of
`Q_(p)` for the given redox reaction is
`1.0* 10^(-8)`

Step-by-step explanation:

Redox reaction with states of species:

`14H^(+)(aq.)+Cr_(2)O_(7)^(2-)(aq.)+6Cl^(-)(aq.)\rightarrow 2Cr^(3+)(aq.)+3Cl_(2)(g)+7H_(2)O(l)`

Reaction quotient for this redox reaction:

`Q_(p)=\frac{[Cr^(3+)]^(2).P_{Cl_(2)}^(3)}{[H^(+)]^(14).[Cr_(2)O_(7)^(2-)].[Cl^(-)]^(6)}`

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of
`H_(2)O` is taken as 1 due to the fact that
`H_(2)O` is a pure liquid.

`pH=-log[H^(+)]`

So,
`[H^(+)]=10^(-pH)`

Plug in all the given values in the equation of
`Q_(p)`:

`Q_(p)=((0.10)^(2)* (0.010)^(3))/((10^(-0.0))^(14)* (1.0)* (1.0)^(6))=1.0* 10^(-8)`