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Calculate the equilibrium constant (K) for the electrochemical reaction: Mg(s) + 2 Ag+(aq) → Mg2+(aq) + 2 Ag(s) Given that the standard reduction potentials for Ag+ and Mg2+ are +0.80 V and -2.37 V, respectively. The concentration of Ag+ is 0.020 M and the concentration of Mg2+ is 0.080 M.

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To calculate the equilibrium constant  K  for the given electrochemical reaction, we first need to find the standard cell potential  E  for the reaction. The standard cell potential can be calculated using the standard reduction potentials of the two half-reactions:E cell  = E cathode  - E anode In this case, the reduction of Ag+ to Ag is the cathode half-reaction, and the oxidation of Mg to Mg2+ is the anode half-reaction. E cell  =  +0.80 V  -  -2.37 V  = 3.17 VNow, we can use the Nernst equation to relate the standard cell potential to the equilibrium constant  K :E cell  =  RT/nF  * ln K Where:- R is the gas constant  8.314 J/molK - T is the temperature in Kelvin  assuming 298 K, which is approximately 25C - n is the number of electrons transferred in the reaction  2 in this case, as 2 electrons are transferred from Mg to 2 Ag+ - F is the Faraday constant  96,485 C/mol Rearranging the equation to solve for K:K = exp nF * E cell  / RT Plugging in the values:K = exp 2 * 96485 * 3.17 /  8.314 * 298  K  1.28  10^32So, the equilibrium constant  K  for the given electrochemical reaction is approximately 1.28  10^32.

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