A battery is constructed of Cr and $$\text{Na}_2\text{Cr}_2\text{O}_7$$. The unbalanced chemical equation when such a battery discharges is following: $$$\text{Na}_2\text{Cr}_2\text{O}_7 + \text{Cr} + \text{H}^+ \rightarrow \text{Cr}^{3+} + \text{H}_2\text{O} + \text{Na}^+$$$ If one Faraday of electricity is passed through the battery during the charging, the number of moles of $$\text{Cr}^{3+}$$ removed from the solution is

Step 1: Understand the Charging Process

The problem provides an unbalanced equation for the discharge of the battery. During charging, the exact reverse reaction takes place. The $$\text{Cr}^{3+}$$ ions present in the solution will be consumed (removed) as they undergo an oxidation or reduction process back to their initial states.

Step 2: Identify Oxidation States and Split the Redox Reactions

Let's find the change in oxidation states for the chromium species during discharge to understand how they behave:

• In $$\text{Na}_2\text{Cr}_2\text{O}_7$$, the oxidation state of Chromium is $$+6$$.
• $$\text{Cr}$$ is in its elemental state, which is $$0$$.
• Both react to form $$\text{Cr}^{3+}$$ (oxidation state $$+3$$).

Therefore, the total process involves two distinct chromium half-reactions during discharge:

1. Oxidation half-reaction: $$\text{Cr} \rightarrow \text{Cr}^{3+} + 3e^-$$
2. Reduction half-reaction: $$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$

Step 3: Equalize Electrons to Balance the Total Redox Reaction

To obtain the overall balanced cell reaction for discharging, we multiply the oxidation half-reaction by 2 so that the number of transferred electrons ($$6e^-$$) matches:

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

Adding these together gives the total balanced discharge reaction:

$$\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{Cr} + 14\text{H}^+ \rightarrow 4\text{Cr}^{3+} + 7\text{H}_2\text{O} + 2\text{Na}^+$$

Step 4: Calculate Moles of $$\text{Cr}^{3+}$$ per Faraday During Charging

From our balanced configuration, we can establish the relationship between the moles of $$\text{Cr}^{3+}$$ and the quantity of electricity (electrons) passed:

• The reaction involves a total transfer of $$6 \text{ moles of electrons}$$ ($$6\text{ Faradays}$$ of electricity).
• According to the stoichiometry of the reaction, a flow of $$6\text{ F}$$ produces or removes $$4\text{ moles of }\text{Cr}^{3+}$$.

Using a simple unitary method for $$1\text{ Faraday}$$ of electricity passed during charging:

$$\text{Moles of }\text{Cr}^{3+}\text{ removed} = \frac{4\text{ moles of }\text{Cr}^{3+}}{6\text{ F}} \times 1\text{ F} = \frac{2}{3}\text{ mole}$$

Conclusion:

Passing one Faraday of electricity through the system will result in the consumption of exactly $$\frac{2}{3}$$ mole of $$\text{Cr}^{3+}$$ from the solution environment.

Answer: Option D — $$\frac{2}{3}$$