ANSWER ONLY 2. One beaker contains a solution 0.0200 M KMnO_4, 0.00500 F MnSO_4

Question

ANSWER ONLY 2.

One beaker contains a solution 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

In the cathode --> reduction occurs

so

the reaction must likely to occur:

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

therefore

Fe goes from +2 to Fe+3

then, this is reducing

then

MnO4- goes from +7 to MnSO4 --> +2

the reaction

MnO4- + 5 e- --> MnSO4

but the full equation mus tinclude the acidic presence

so

MnO4 + 8 H+ + 5 e Mn2+ + 4 H2O

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