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What is the number of electrons transferred during the redox reaction between copper metal and silver ion, if a solution containing 0.05 moles of copper metal is used with 0.1 moles of silver ion?

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The redox reaction between copper metal and silver ion can be represented as follows:Cu s  + 2Ag aq   Cu aq  + 2Ag s In this reaction, copper metal  Cu  loses 2 electrons to become Cu, while each silver ion  Ag  gains 1 electron to become Ag. Since there are 0.05 moles of copper metal and 0.1 moles of silver ions, the reaction will proceed as follows:0.05 moles Cu   2 moles Ag / 1 mole Cu  = 0.1 moles AgThis means that all the copper metal and silver ions will react completely.Now, we can calculate the number of electrons transferred:For copper:0.05 moles Cu   2 electrons / 1 mole Cu  = 0.1 moles of electronsFor silver:0.1 moles Ag   1 electron / 1 mole Ag  = 0.1 moles of electronsSince both copper and silver have the same number of moles of electrons transferred, the total number of electrons transferred during the redox reaction is:0.1 moles of electrons   6.022 x 10 electrons / 1 mole of electrons   6.022 x 10 electrons

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