The quantity of electricity in Faraday needed to reduce 1 mol of Cr2O _7^2 - to Cr^3+ is \underline\hspace2cm .

Electrochemistry question

2022 · 28 Jun · Shift 1 · Q20

2022 · 28 Jun · Shift 1 · Q20

View written solutionFree

Correct answer: 6

We need the number of moles of electrons required to reduce $1$ mol of $\text{Cr}_2\text{O}_7^{2-}$ to $\text{Cr}^{3+}$ .
Write the reduction half-reaction in acidic medium:

\text{Cr}_2\text{O}_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O

From the balanced half-reaction, $1$ mol of $\text{Cr}_2\text{O}_7^{2-}$ requires $6$ mol of electrons.
Since $1$ Faraday corresponds to $1$ mol of electrons, the quantity of electricity required is:

6 \text{ Faraday}

\boxed{6}

2022 · 28 Jun · Shift 1 · Q20

2022 · 28 Jun · Shift 1 · Q20

View written solutionFree

Correct answer: 6

We need the number of moles of electrons required to reduce $1$ mol of $\text{Cr}_2\text{O}_7^{2-}$ to $\text{Cr}^{3+}$ .
Write the reduction half-reaction in acidic medium:

\text{Cr}_2\text{O}_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O

From the balanced half-reaction, $1$ mol of $\text{Cr}_2\text{O}_7^{2-}$ requires $6$ mol of electrons.
Since $1$ Faraday corresponds to $1$ mol of electrons, the quantity of electricity required is:

6 \text{ Faraday}

\boxed{6}