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Calculate the number of electrons transferred during the following redox reaction: Cu(s) + 2Ag+  -->  Cu2+ + 2Ag(s)

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To calculate the number of electrons transferred during the redox reaction, we need to determine the change in oxidation states for each element involved in the reaction.In the given reaction:Cu s  + 2Ag+ --> Cu2+ + 2Ag s The oxidation states of the elements are as follows:Cu: 0  in its elemental form Ag+: +1  as a cation Cu2+: +2  as a cation Ag: 0  in its elemental form Now, let's determine the change in oxidation states:For Cu: It goes from an oxidation state of 0 to +2, which means it loses 2 electrons. This is the oxidation half-reaction: Cu --> Cu2+ + 2e-For Ag+: It goes from an oxidation state of +1 to 0, which means it gains 1 electron. Since there are 2 moles of Ag+ in the reaction, a total of 2 electrons are gained. This is the reduction half-reaction: 2Ag+ + 2e- --> 2AgIn this redox reaction, a total of 2 electrons are transferred from Cu to 2Ag+.

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