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Chapter 18: Problem 130

In the electrolysis of a sodium chloride solution, what volume of \(\mathrm{H}_{2}(g)\) is produced in the same time it takes to produce 257 \(\mathrm{L} \mathrm{Cl}_{2}(g),\) with both volumes measured at $50 .^{\circ} \mathrm{C}\( and 2.50 \)\mathrm{atm}$ ?

Short Answer

Expert verified
Under the given conditions of \(50 .^{\circ}\mathrm{C}\) and \(2.50\,\mathrm{atm}\), the volume of hydrogen gas produced in the same time as 257 L of chlorine gas is also 257 L.

Step by step solution

01

Write down the balanced redox reactions and overall equation

The full balanced equation for the electrolysis of aqueous sodium chloride (NaCl) is: \(2 \mathrm{Na}^{+} + 2 \mathrm{Cl}^{-}(aq) \rightarrow 2 \mathrm{Na}(s) + \mathrm{Cl}_{2}(g)\) (at the anode) \(2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2 \mathrm{OH}^{-} + \mathrm{H}_{2}(g)\) (at the cathode) Summing these reactions: \(2 \mathrm{Na}^{+} + 2 \mathrm{Cl}^{-} + 2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2 \mathrm{Na}(s) + \mathrm{Cl}_{2}(g) + 2 \mathrm{OH}^{-} + \mathrm{H}_{2}(g)\)
02

Determine the stoichiometry of hydrogen gas and chlorine gas produced

From the balanced equation, we can see that 1 mole of hydrogen gas is produced for every mole of chlorine gas produced: \(\frac{\mathrm{moles\, of\, H}_{2}}{\mathrm{moles\, of\, Cl}_{2}} = \frac{1}{1}\) If we're given the volume of \(\mathrm{Cl}_{2}\), we can use this ratio to determine the volume of \(\mathrm{H}_{2}\): \(\frac{\mathrm{volume \,of\, H}_{2}}{\mathrm{ volume\, of\, Cl}_{2}} = \frac{1}{1}\)
03

Using the stoichiometry, find the volume of hydrogen gas

We are given the volume of chlorine gas as 257 L. Using the stoichiometry ratio from Step 2: \(\mathrm{volume\, of\, H}_{2} = \mathrm{volume\, of\, Cl}_{2} = 257\,L\)
04

Verify the given conditions are the same for both gases

To ensure the volumes of hydrogen and chlorine gases are directly comparable, verify that they are both measured at the same temperature and pressure conditions. Both gases are measured at \(50 .^{\circ}\mathrm{C}\) and \(2.50\, \mathrm{atm}\), so the volume of hydrogen gas would also be equal to the volume of chlorine gas under these same conditions.
05

Conclude the answer

Under the given conditions of \(50 .^{\circ}\mathrm{C}\) and \(2.50\, \mathrm{atm}\), the volume of hydrogen gas produced in the same time as 257 L of chlorine gas is also 257 L.

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Most popular questions from this chapter

Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. $\mathrm{Cr}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{NO}(g)$ b. $\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Ce}^{4+}(a q) \rightarrow \mathrm{CO}_{2}(a q)+\mathrm{Ce}^{3+}(a q)$ c. $\mathrm{SO}_{3}^{2-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Mn}^{2+}(a q)$

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