The lead electrode is immersed in a solution containing 0.01 m sodium iodide, saturated with lead iodide; paired with n.c.e, it serves as an anode. Calculate the PR (PbI2) if the system potential is 0.54 V. E0Pb2 / Pb0=-0.126 V. E (n.c.e) = 0.247 V
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