MISSED THIS? Read Section 20.6 (Pages 917 - 922) . A voltaic cell employs the following redox reaction: Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq) Calculate the cell potential at 25 ∘C under each of the following conditions. Part A standard conditions Ecell = nothing V Request Answer Part B [Sn2+]= 1.17×10−2 M ; [Mn2+]= 1.74 M . Express your answer using two significant figures. Ecell = nothing V Request Answer Part C [Sn2+]= 1.74 M ; [Mn2+]= 1.17×10−2 M .

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

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