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Using a table of standard reduction potentials, calculate the X-cell potential for the reduction of aqueous lead(10) ions to
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1939-3: 1.5x109 m To 270 = 300.15 K Pb²+ (aq) + 2 € 7 Pb (s) Epp2r/pn = – 9.13 v Imo. of e transferred, mz 2 from E = NernstNernst equation is used to calculate the half cell potential of this half cell.

DelG = nFE is used to calculate the free energy change in this cell.

** Ignore if there is any calculations mistake, but the process is spot on.

** Feel free to comment in any kind of doubt.

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