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October 26, 2014

October 26, 2014

Posted by **sabrina** on Friday, November 30, 2012 at 11:23pm.

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)

____________

(b) Mg(s) | Mg2+(0.070) M || [Al(OH)4]–(0.30 M), OH–(0.040 M) | Al(s)

- CHEMISTRY -
**DrBob222**, Friday, November 30, 2012 at 11:48pma) 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.

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