For Cr_2O_7^{2-} + 14H^{+} + 6e^- longrightar | vedclass

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(A) According to the Nernst equation at $298 \ K$:$E = E^{\circ} - \frac{0.0591}{n} \log Q$Here,$n = 6$. The reaction quotient $Q$ is given by $Q = \f

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$2$

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(A) According to the Nernst equation at $298 \ K$:$E = E^{\circ} - \frac{0.0591}{n} \log Q$Here,$n = 6$. The reaction quotient $Q$ is given by $Q = \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}] [H^{+}]^{14}}$.Substituting the values:$1.067 = 1.33 - \frac{0.0591}{6} \log \left( \frac{(1.5 	imes 10^{-3})^2}{(4.5 	imes 10^{-3}) [H^{+}]^{14}} ight)$$1.33 - 1.067 = \frac{0.0591}{6} \log \left( \frac{2.25 	imes 10^{-6}}{4.5 	imes 10^{-3} [H^{+}]^{14}} ight)$$0.263 = 0.00985 \log \left( \frac{0.5 	imes 10^{-3}}{[H^{+}]^{14}} ight)$$26.7 = \log \left( \frac{5 	imes 10^{-4}}{[H^{+}]^{14}} ight)$$10^{26.7} = \frac{5 	imes 10^{-4}}{[H^{+}]^{14}}$Solving for $[H^{+}]$,we find $[H^{+}] = 10^{-2} \ M$.$pH = -\log [H^{+}] = -\log(10^{-2}) = 2$.

For $Cr_2O_7^{2-} + 14H^{+} + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$,$E^0 = 1.33 \ V$. Given $[Cr_2O_7^{2-}] = 4.5 \ mmol$,$[Cr^{3+}] = 1.5 \ mmol$ and $E = 1.067 \ V$,calculate the $pH$ of the solution.

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