One beaker contains a solution of 0.0200 M KMnO_4, 0.00500 F MnSO_4 and 0.500 F

Question

One beaker contains a solution of 0.0200 M KMnO_4, 0.00500 F MnSO_4 and 0.500 F H_2SO_4. A second beaker contains 0.150 F FeSO_4)and 0.00150 F Fe_2(SO_4)_3. Pt wires are placed in each and connected via wires to a voltmeter A salt bridge is used to connect the beakers and the second beaker is connected as the anodic half cell. Given temperature of 298 K and using information in appendix H and other places in the book, answer the following. What is the net ionic half reaction at the anode (exclude spectator ions) and its E degree value: What is the net ionic half reaction at the cathode (exclude spectator ions) and its E degree value: What is the E degree value for the cell (show equation)? Give the balanced cell reaction, do not include spectator ions:

Explanation / Answer

4.a

balance equation

2 KMNO4 + 10 FeSO4 + 8 H2SO4 = K2SO4 + 2 MNSO4 + 5 Fe2(SO4)3 + 8 H2O

explanation

Mn goes from +7 to +2, => -5
Fe goes from +2 to +3, => +1

You need 5 Fe atoms per Mn atom.

KMnO4 + 5FeSO4 + ?H2SO4 2.5 Fe2(SO4)3 + 0.5K2SO4 + MnSO4 + ?H2O

Since fractions look silly in equations, multiply all by 2.

2KMnO4 + 10FeSO4 + ?H2SO4 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + ?H2O

Now balance the rest.

2KMnO4 + 10FeSO4 + 8H2SO4 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O

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