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Answer :
The subject area is Chemistry. The provided information refers to an electrochemical cell or a battery setup with Sn as the anode and Cu as the cathode. We can use the Nernst Equation to calculate the initial E (electromotive force) given the concentrations of Sn. However, calculating a new E value after running the battery for a certain period requires additional information.
The subject of question refers to a battery setup with metals of Sn and Cu, indicative of an electrochemistry problem. The E for a battery is the electromotive force, or potential difference, which drives the current flow in the battery.
According to the standard reduction potentials table, Sn+2 is reduced to Sn(s) at -0.14V and Cu+2 is reduced to Cu(s) at +0.34V.
In an electrochemical cell, typically the element that is oxidized acts as the anode (losing electrons) and the one reduced acts as the cathode (gaining electrons). Here, Sn goes from elemental form to ion form, it loses two electrons in the process (is oxidized) and therefore, is the anode. On the other hand, Cu accepts two electrons (is reduced), making it the cathode. So, Sn is oxidized at the anode and Cu is reduced at the cathode.
The E for this battery can be calculated using the equation for the Nernst Equation:
Since we are given the concentrations of Sn+2 and Cu+2, we can use these values to find Q (the reaction quotient), E° (standard cell potential) can be obtained from standard reduction potentials and n is the number of electron transfers during the reaction.
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