One Faraday of electricity liberates $x \times 10^{-1}$ gram atom of copper from copper sulphate. $x$ is ________.

<p>To find the value of <i>x</i> when one Faraday of electricity liberates <i>x</i> times $10^{-1}$ gram atom of copper from copper sulphate, we need to understand how electricity interacts with copper ions in solution. The key reaction is:</p><p>$\mathrm{Cu}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}$</p><p>This shows that copper ions (Cu²⁺) gain two electrons (e^-) to become copper metal (Cu). From electrochemistry, we know that:</p><ul><li>2 Faradays of electricity are required to deposit 1 mole (or 1 gram atom) of copper.</li><li>Therefore, 1 Faraday will deposit half of that amount, which is 0.5 mole, or in other terms, 0.5 gram atom of copper.</li><li>Expressing 0.5 in the form of <i>x</i> times $10^{-1}$ gives us $5 \times 10^{-1}$.</li></ul><p>This means <i>x</i> equals 5.</p>