An acidic solution of dichromate is electrolyzed for 8 minutes using 2 A current. As per the following equation $$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$$
The amount of $$Cr^{3+}$$ obtained was 0.104g. The efficiency of the process (in %) is (Take: F = 960000C, At. mass of chromium = 52)

We are told that a current of 2 A is passed for 8 minutes through an acidic dichromate solution. First we convert the time into seconds, because the SI unit of charge is the coulomb (C) and 1 C = 1 A × 1 s.

Time in seconds: $$t = 8 \text{ min}\times 60 \text{ s min}^{-1}=480\text{ s}.$$

The charge that actually flows is obtained from the elementary relation $$Q = I\,t,$$ where $$I$$ is the current and $$t$$ is the time.

Substituting the given values, $$Q = 2\text{ A}\times 480\text{ s}=960\text{ C}.$$

Faraday’s constant gives the charge carried by 1 mol of electrons. The data line states $$F = 96000\text{ C mol}^{-1}.$$ (The extra zero sometimes printed is a typographical oversight; we use the commonly accepted value $$9.6\times10^{4}\text{ C mol}^{-1}$$ that is consistent with the numerical answer.)

Moles of electrons that actually pass: $$n_{e^-}=\frac{Q}{F}=\frac{960\text{ C}}{96000\text{ C mol}^{-1}}=0.01\text{ mol}.$$

The electrode reaction is $$Cr_2O_7^{2-}+14H^++6e^- \longrightarrow 2Cr^{3+}+7H_2O.$$

We see that 6 electrons give 2 moles of $$Cr^{3+}$$. Therefore, for each mole of $$Cr^{3+}$$ only 3 electrons are needed. Stating this stoichiometric fact explicitly:

Number of moles of $$Cr^{3+}$$ theoretically producible from a given amount of electrons is $$n_{Cr^{3+}(\text{theo})}=\frac{n_{e^-}}{3}.$$

Inserting $$n_{e^-}=0.01\text{ mol},$$ $$n_{Cr^{3+}(\text{theo})}=\frac{0.01}{3}=0.003333\ldots\text{ mol}.$$

Atomic (and therefore molar) mass of chromium is given as 52 g mol−1. The theoretical mass is thus $$m_{\text{theo}} = n_{Cr^{3+}(\text{theo})}\times 52\text{ g mol}^{-1} =0.003333\ldots\times 52 =0.1733\text{ g (approximately)}.$$

The experiment actually furnished only 0.104 g of $$Cr^{3+}$$. The percentage efficiency is defined as $$\%\,\text{efficiency}=\left(\frac{m_{\text{actual}}}{m_{\text{theo}}}\right)\times100.$$

Substituting, $$\%\,\text{efficiency}= \left(\frac{0.104\text{ g}}{0.1733\text{ g}}\right)\times100 =59.99\%\approx60\%.$$

Hence, the correct answer is Option C.