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  • The quantity of elctricity in Faraday needed to reduce <img alt="1 \text { mol of } \mathrm{Cr}_{2} \mathrm{O}_{7}^{-2} \text { to } \mathrm{Cr}^{3+}" src="https://entrancecorner.oncodecogs.co

The quantity of elctricity in Faraday needed to reduce 1 \text { mol of } \mathrm{Cr}_{2} \mathrm{O}_{7}^{-2} \text { to } \mathrm{Cr}^{3+} is _______.

Option: 1

6


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{given,}
        \begin{aligned} &\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{2Cr^{3+}} \end{aligned}

        \begin{aligned} \mathrm{Cr}={+6} \: \: \: \: \: \mathrm{Cr=+3 }\end{aligned}

        \mathrm{n \text {-factor }=2(6-3)=6}

So, 6 moles of electrons are required to reduce 1 mole of \mathrm{Cr_2O_7^{2-}}

\mathrm{1 \: mol \: of\, electron =1\: Faraday. }

\mathrm{6 \text { mol of } e^{\ominus }=6 \text { Faraday. }}

Answer is 6

Posted by

Shailly goel

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