1. Potassium dichromate in acidic solution is used to titrate a solution of iron

Question

1. Potassium dichromate in acidic solution is used to titrate a solution of iron(II) ions, with which it reacts according to Cr2O7 2- (aq) + 6Fe2+ (aq) + 14H3O+ (aq) 2Cr3+ (aq) + 6Fe3+ (aq) + 21H2O (l) A potassium dichromate solution is prepared by dissolving 5.134 g of K2Cr2O7 in water and diluting to a total volume of 1.000 L. A total of 34.26 mL of this solution is required to reach the end point in a titration of a 500.0-mL sample containing Fe2+ (aq). Determine the concentration of Fe2+ in the original solution. From

2. Calculate the potential E° of a lead-acid cell if all reactants and products are in their standard states. What will be the voltage if six such cells are connected in a series?

Explanation / Answer

The reaction is given as

Cr2O7 2- (aq) + 6Fe2+ (aq) + 14H3O+ (aq) -----> 2Cr3+ (aq) + 6Fe3+ (aq) + 21H2O (l)

So 1 mole of K2Cr2O7 reacts with 6 moles of Fe2+

5.134 g of K2Cr2O7 in water and diluting to a total volume of 1.000 L

Means 5.134/294g/mol = 0.0174 moles in 1 L is 0.0174 M

34.26 mL of this solution is required to reach the end point in a titration of a 500.0-mL sample containing Fe2+ (aq).

V1N1 = V2N2

34.26mL x 0.0174 = 0.596 mmol of K2Cr2O7

So iron will be 0.596 x 6 = 3.576 mmol

in 500 mL it is 3.576/500 = 0.00715 M in Fe2+

In a lead acid cell the two half reactions are

Negative plate reaction Pb(s) + SO42-(aq) ---> PbSO4(s) + 2 e-   E = +0.356Positive plate reactionPbO2(s) + SO42-(aq) + 4 H+(aq) + 2 e-    -----> PbSO4(s) + 2 H2O E = +1.685

Total reaction is

Pb(s) + 2SO42-(aq) + PbO2(s)  + 4 H+(aq) ---> 2PbSO4(s) + 2H2O(l) Eo = 1.685 + 0.356 = 2.041 V

If 6 are connected in series you will get 2.041 x 6 = 12.246 V

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