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  • Given that the pH of the reaction is 4 and at 298 K 1 litre solution containing 10 m mole of <img alt="\mathrm{Cr}_2 \mathrm{O}_7^{2-}" src="https://entrancecorner.oncodecogs.com/gif.late

Given that the pH of the reaction is 4 and at 298 K 1 litre solution containing 10 m mole of \mathrm{Cr}_2 \mathrm{O}_7^{2-} and 100m mol of \mathrm{Cr}^{3+}. Calculate the potential for the half cell rxm.

\mathrm{\mathrm{C}_2 \mathrm{O}_7^{2-} \rightarrow \mathrm{Cr}^{3+}\left\{E^0=1.330 \mathrm{~V} \text { and } \frac{2.303 \mathrm{RT}}{\mathrm{F}}\right. is ~0.059 V\}}

Option: 1

23 \times 10^{-6} \mathrm{~V} \\
 


Option: 2

56 \times 10^{-1} \mathrm{~V} \\


Option: 3

84 \times 10^{-3} \mathrm{~V} \\


Option: 4

77 \times 10^{-2} \mathrm{~V} \\


Answers (1)

best_answer

\mathrm{CrO}_7^{2-}+14 \mathrm{H}^{+}+6 e^{-} \rightarrow \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}

\mathrm{E R=E^0 R - \frac{0.059}{6} \log \frac{\left|\mathrm{Cr}^{3+}\right|^{2}}{\mathrm{Cr}_2 \mathrm{O}_7^{2-}\left[\mathrm{H}^{+}\right]^{14}}}

\mathrm{E R=1.33-0.01 \log \frac{\left(10^{-1}\right)^2}{\left(10^{-2}\right)\left(10^{-4}\right)^{14}}}

\mathrm{ E R=1.33-0.01 \log (10)^{56} \\ }

\mathrm{ E R=1.33-0.01(56) \log 10 \\ }

\mathrm{ E R=0.77 \\ }

\mathrm{ E R=77 \times 10^{-2} \mathrm{~V} }

Posted by

Deependra Verma

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