KCET · Chemistry · Electrochemistry
The amount of current in Faraday is required for the reduction of \( 1 \mathrm{~mol} \) of \( \mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-} \) ions to
\( C r^{3+} \) is,
- A \( 1 \mathrm{~F} \)
- B - \( 2 \) F
- C \( 6 \mathrm{~F} \)
- D F
Answer & Solution
Correct Answer
(C) \( 6 \mathrm{~F} \)
Step-by-step Solution
Detailed explanation
The balance reaction is,
\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+4 \mathrm{H}^{+}+6 e^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\)
From the reaction, \(1 \mathrm{~mol}\) of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) requires \(6 \mathrm{~F}\) to convert into \(\mathrm{Cr}^{3+}\) ion.
\(\mathrm{Q}=\mathrm{nF}\) (where, \(\mathrm{n}=\) no. of \(e^{-}\))
\(Q=6 \mathrm{~F}\)
\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+4 \mathrm{H}^{+}+6 e^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\)
From the reaction, \(1 \mathrm{~mol}\) of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) requires \(6 \mathrm{~F}\) to convert into \(\mathrm{Cr}^{3+}\) ion.
\(\mathrm{Q}=\mathrm{nF}\) (where, \(\mathrm{n}=\) no. of \(e^{-}\))
\(Q=6 \mathrm{~F}\)
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