Answer :
That Sn will be the anode, and Cu will be the cathode in the battery. After 12 hours of operation at 4.2 amps, the updated E value cannot be accurately determined due to missing initial reduction potentials.
In batteries, oxidation occurs at the anode and reduction at the cathode. Sn+2/Sn0 is the pair undergoing oxidation or reduction. Sn+2 will gain electrons, reducing to Sn0, implying that Sn acts as the anode. Similarly, Cu+1/Cu0 is the pair where Cu+1 loses an electron to become Cu0, meaning Cu is the cathode.
When the battery operates at 4.2 amps for 12 hours, this results in a change in the concentrations of the chemical species in the cell, affecting its E value (cell potential). However, without the initial reduction potentials for these reactions, it's not feasible to calculate the precise change in E.
The identification of the anode and cathode materials can be done using oxidation and reduction rules, but the calculation of the new E value requires more data.
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