What is the CORRECT expression of potential (versus SHE) at the equivalence point(EEP) for the titration of ceric ion (Ce4+) with plumbous ion (Pb2+)?Ce4+ +e- -->Ce3+Pb4+ + 2e-Pb2+-->O EEP = [(E°Ce + 2E°pb)/3] – (0.0592/3) log ([Ce³"][Pb4+/[Ce++][Pb2+])O EEP = (E°Ce + E°pb)/2O EEP = (E°Ce + 2E°pb)/3O EEP = (E°Ce + E°pb) – 0.0592 log ([Ce³+][Pb2+]/[Ce4+][Pb4+j)

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What is the CORRECT expression of potential (versus SHE) at the equivalence point EEP) for the titration of ceric ion (Ce4+) with plumbous ion (Pb2+)? Ce4+ +e- --> Ce³+ Pb4+ + 2e- --> Pb2+ 2 EEP = [(E°Ce + 2E°pb)/3] – (0.0592/3) log ([Ce³+][Pb4+/[Ce4+][Pb²+j) O EEP = (E°Ce + E°pb)/2 O EEP = (E°Ce + 2E°pb)/3 O EEP = (E°Ce + E°pb) – 0.0592 log ([Ce³+][Pb²+J/[Ce4+][Pb4+j)

What is the CORRECT expression of potential (versus SHE) at the equivalence point EEP) for the titration of ceric ion (Ce4+) with plumbous ion (Pb2+)? Ce4+ +e- --> Ce³+ Pb4+ + 2e- --> Pb2+ 2 EEP = [(E°Ce + 2E°pb)/3] – (0.0592/3) log ([Ce³+][Pb4+/[Ce4+][Pb²+j) O EEP = (E°Ce + E°pb)/2 O EEP = (E°Ce + 2E°pb)/3 O EEP = (E°Ce + E°pb) – 0.0592 log ([Ce³+][Pb²+J/[Ce4+][Pb4+j)

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.20QAP

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