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Chapter 19: Problem 58

A quantity of \(0.300 \mathrm{~g}\) of copper was deposited from a \(\mathrm{CuSO}_{4}\) solution by passing a current of \(3.00 \mathrm{~A}\) through the solution for 304 s. Calculate the value of the faraday constant.

Short Answer

Expert verified
The value of Faraday's Constant is approximately 96500 A*s/mol.

Step by step solution

01

Converting Units

First, convert the values into their proper units suitable for calculations. 1. Convert the mass m of copper to kg (since 1g = 0.001kg): m = 0.300g * 0.001kg/g = 0.0003 kg. 2. Ensure that current is in amperes: I = 3 A. 3. Ensure that time is in seconds: t = 304 s. 4. Convert the molar mass of copper to kg/mol (since 1g = 0.001kg): M = 63.546 g/mol * 0.001 kg/g = 0.063546 kg/mol. 5. Number of electrons involved in the reduction of copper ions: n = 2.
02

Calculate Faraday's Constant

Plug the values into the Faraday's Constant formula: F = \( \frac{mIt}{Mn} \) = \( \frac{0.0003 kg * 3A * 304s}{0.063546 kg/mol * 2} \) = 96487.799 A*s/mol, which is approximately = 96500 A*s/mol.

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Most popular questions from this chapter

The half-reaction at an electrode is $$ \mathrm{Mg}^{2+}(\text { molten })+2 e^{-} \longrightarrow \mathrm{Mg}(s) $$ Calculate the number of grams of magnesium that can be produced by supplying \(1.00 \mathrm{~F}\) to the electrode.

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