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Electrochemistry question

2020 · 3 Sep · Shift 2 · Q21

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Electrochemistry
2020 · 3 Sep · Shift 2 · Q21

Electrochemistry question

2020 · 3 Sep · Shift 2 · Q21

JEE MainChemistryElectrochemistryNumerical+4 / −1

An acidic solution of dichromate is electrolyzed for 8 minutes using 2A current. As per the following equation

Cr_2O_7^{2-}

+

14H^+

+ 6e–

\to

2Cr^{3+}

+

7H_2O

The amount of

Cr^{3+}

obtained was 0.104 g. The efficiency of the process(in%) is (Take : F = 96000 C, At. mass of chromium = 52)

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 60

Given data

Current, $I = 2\,\text{A}$
Time, $t = 8\,\text{min} = 480\,\text{s}$
Charge passed, $Q = It = 2 \times 480 = 960\,\text{C}$
Cathode reaction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$
Faraday constant, $F = 96000\,\text{C}$
Atomic mass of chromium $= 52$
Actual mass of $Cr^{3+}$ obtained $= 0.104\,\text{g}$

Find theoretical amount of $Cr^{3+}$ produced

From the reaction:

$6$ moles of electrons produce $2$ moles of $Cr^{3+}$
So, $3$ moles of electrons produce $1$ mole of $Cr^{3+}$

Thus, equivalent electrons needed for $1$ mole of $Cr^{3+}$ is $3$ .

Using Faraday's law:

$\text{moles of } Cr^{3+} = \frac{Q}{3F} = \frac{960}{3 \times 96000} = \frac{1}{300}$

So theoretical mass of $Cr^{3+}$ is

$m_{\text{theoretical}} = \frac{1}{300} \times 52 = 0.1733\,\text{g}$

Calculate current efficiency

$\text{Efficiency} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100$

$= \frac{0.104}{0.1733} \times 100$

$= 60\%$

Final answer

The efficiency of the process is:

$\boxed{60}$

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