Faraday’s second law of electrolysis: "The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte."
At the anode: 2Cl− → Cl2(g) + 2e− Chlorine gas (Cl2) is evolved, and the solution near the anode becomes more acidic (pH decreases).
At the cathode: 2H2O + 2e− → H2(g) + 2OH− Hydrogen gas (H2) is evolved, and the solution near the cathode becomes more alkaline (pH increases).
If E$_{cell}$ of the following reaction is x $\times$ 10$^{-1}$. Find x
\(\text{Pt/ HSnO$_2$ / Sn(OH)$_6^{2-}$, OH$^-$ / Bi$_2$O$_3$ / Bi / Pt}\)
\(\text{[Reaction Quotient, Q = 10$^6$]}\)
Given \( E^o_{\text{[Sn(OH)$_3$]}} \) = -0.90 V, \( E^o_{\text{Bi$_2$O$_3$ / Bi}} \) = -0.44 V

From the following information, calculate opening and closing inventory :