What is $\varepsilon$ of the following cell reaction at 25°C? $\varepsilon$°_cell = 0.460 V. Cu(s)| Cu²⁺(0.014 M)|| Ag⁺(0.20 M)| Ag(s)

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The Nernst equation is used to calculate the cell potential under non-standard conditions. The equation is: E = E° - (RT/nF) * ln(Q), where E is the cell potential under non-standard conditions, E° is the standard cell potential, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (298K for 25°C), n is the number of moles of electrons transferred in the reaction, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient. Given E° = 0.460 V, and the concentrations of Cu^2+ and Ag^+, we can calculate Q = [Ag^+]/[Cu^2+] = 0.20/0.014. Since the reaction involves the transfer of 2 moles of electrons (Cu to Cu^2+ and Ag^+ to Ag), n = 2. Plugging these values into the Nernst equation allows us to calculate the cell potential under the given conditions, which results in approximately 0.473 V.

What Is ε\varepsilonε Of the Following Cell Reaction at 25°C ε\varepsilonε °cell = 0

What Is $\varepsilon$ Of the Following Cell Reaction at 25°C $\varepsilon$ °_cell = 0