a) Mn(s) | Mn2+(0.11) M || Cr3+(0.24 M), Cr2+(0.11 M) | Pt(s)
1. Calculate4 the Mn^2+ + e ==> Mn couple.
E = Eo - 0.0592/n)*log(red/(ox)
E = Eo-(0.0592/n)*log[Mn(s)]/(Mn^2+)
E = Eo = (0.0592/2)*log[(1)/(0.11)]
You will need ot look up Eo for this REDUCTION, calculate E for the half cell, then change the sign since this is an oxidation.
Do the same for the Cr^3+ + e ==> Cr^2+
Then EMn as oxidation + ECr as redn = Ecell.
Use the Nernst equation and data from Appendix D to calculate Ecell for each of the following cells.
Enter your answer with 3 significant digits.
Enter scientific notation as 1.23E4.
(a) Mn(s) | Mn2+(0.11) M || Cr3+(0.24 M), Cr2+(0.11 M) | Pt(s)
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(b) Mg(s) | Mg2+(0.070) M || [Al(OH)4]–(0.30 M), OH–(0.040 M) | Al(s)
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