At 298 K,a 1 L solution containing 10 mmol | vedclass

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Question

(C) The half-cell reaction is: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O$.Given concentrations: $[Cr_2O_7^{2-}] = 10 \ mmol/L = 0.

Vedclass · Accepted Answer

$917$

Vedclass · Answer

(C) The half-cell reaction is: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} ightarrow 2Cr^{3+} + 7H_2O$.Given concentrations: $[Cr_2O_7^{2-}] = 10 \ mmol/L = 0.01 \ M$,$[Cr^{3+}] = 100 \ mmol/L = 0.1 \ M$,and $pH = 3.0$,so $[H^{+}] = 10^{-3} \ M$.Using the Nernst equation: $E = E^0 - \frac{0.059}{n} \log \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^{+}]^{14}}$.Substituting the values: $E = 1.33 - \frac{0.059}{6} \log \frac{(0.1)^2}{(0.01)(10^{-3})^{14}}$.$E = 1.33 - \frac{0.059}{6} \log \frac{10^{-2}}{10^{-2} 	imes 10^{-42}} = 1.33 - \frac{0.059}{6} \log (10^{42})$.$E = 1.33 - \frac{0.059}{6} 	imes 42 = 1.33 - 0.059 	imes 7 = 1.33 - 0.413 = 0.917 \ V$.Thus,$E = 917 	imes 10^{-3} \ V$,so $x = 917$.

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