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Use the References to access important values if needed for this question. When the Pb2+ concentration is 1.41 M, the observeWhen the Pb2+ concentration is 6.90x10-4 M, the observed cell potential at 298K for an electrochemical cell with the followin

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Pb2+ (aq) + n(s) Pb (w + Mn2+ cor) ...) Reduction half reaction: Pb2+ + 2ē -> Pb(3) Ereq = -0.13V oxidation half reaction: mmPugging all these value into equation (in) 8:314 X 298 1170 - 106 - 2 x 96500 =) 1170 .- 1.06 - 00128 ln & =) 0.0128bn Q = 0here, & cell = 0.5177 (given) using earration (uj) = 67 0.517 = 00 - 8.3/4 X 298 6 x 96500 le Q = 0.517 = 0.61 – 0.00028 ln &

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