As part of your pre$-$lab preparation, BALANCE this redox reaction.

MnO$4$

$+$ Fe$2++$ H

$+->$ Mn$2++$ Fe$3++$ H$2$OMOLARITY OF KMnO$4$

$$ Using the mass of Fe$($NH$4)2($SO$4)26$H$2$O$,$ CALCULATE the moles of Fe$($NH$4)2($SO$4)26$H$2$O and, in turn, the

moles of Fe$2+$

$.$

$*$Remember to include the six H$2$O$$s when calculating the molecular weight.

$$ Use your balanced equation to DETERMINE the moles of KMnO$4$ needed to react with all of the Fe$2+$

$.$

$$ SUBTRACT the final burette reading from the initial burette reading to determine the volume of KMnO$4$ used

in the titration $($to two decimal places$).$

$$ From the moles of KMnO$4$ and the volume in liters, DETERMINE the molarity of the KMnO$4$ solution with

proper significant figures.To determine the normality of your KMnO$4$ solution:

$$ First, BALANCE the half$-$reaction below to determine the number of electrons transferred.

MnO$4$

$+$ H

$++$ e

$->$ Mn$2++$ H$2$O

$$ Based on the molarity of your KMnO$4$ solution and the number of electrons transferred, CALCULATE the

normality of the KMnO$4$ solution.REDOX TITRATION

Name Partner Name$($s$)$

DATA TABLE. Don$$t forget to include units and show your work for processed data!

RAW DATA:

initial KMnO$\_4$ buret reading $0\mathrm{.}02332$L

final KMnO$\_4$ buret reading $0\mathrm{.}06011$L

weight of Fe$($NH$\_4)\_2($SO$\_4)\_26$H$\_2$ O $1\mathrm{.}24$ grams

PROCESSED DATA $($show all work for credit$)$:

moles of Fe$($NH$\_4)\_2($SO$\_4)\_2$ in flask

# of electrons donated PER Fe$($NH$\_4)\_2($SO$\_4)\_2$ unit$/$compound $($from balanced half$-$reaction$)$

TOTAL moles of electrons donated by Fe

TOTAL moles of electrons accepted by KMnO$\_4$

# of electrons absorbed PER KMnO$\_4$ unit

$($from balanced half$-$reaction$)$

moles of KMnO$\_4$ added to flask

volume of KMnO$\_4$ solution added to flask $0\mathrm{.}03679$L

$[$KMnO$\_4]$

QUESTIONS ON BACK$$

Performing redox reactions with manganese is tricky because it has six possible oxidation states in solution. Under basic conditions, the product of this redox reaction is MnO$\_2$ instead of Mn$^(2+).$

$1.$ DETERMINE the NEW manganese half$-$reaction, and WRITE the full, balanced redox reaction.

$2.$ Assuming your RAW DATA remained the same, WHAT WOULD your calculated concentration of potassium permanganate be if the manganese had converted into MnO$\_2$ instead of Mn$^(2+)?$