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

Answer:

Value of $Q_{p}$ for the given redox reaction is $1.0\times 10^{-8}$

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 $Q_{p}$:

$Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}$