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Problem- Ub tge titration of 25.0 mL of 0.1 Fe2+ with 0.100 MCe4+, what will be the cell potential after the addition of 25.0 mL of Ce4+ solution (at the equivalence point)
Ce4+ + e- ---> Ce3+ 1.44 V
Fe3+ + e- ----> Fe2+ 0.77 V
The answer is 1.11 V I just am unaware how to get there.
Provide your answer in scientific notation
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