Use the appropriate standard reduction potentials below to...

Use the appropriate standard reduction potentials below to determine the equilibrium constant at 246 K for the following reaction under acidic conditions. 4H^+(aq) + MnO2(s) + 2Fe^2+(aq) -> Mn^2+(aq) +2Fe^3+(aq) + 2H2O(l) Standard reduction potentials: MnO2(s) + 4H^+(aq) + 2e^- -> Mn^2+(aq) + 2H2O(l) E standard = 1.23 V Fe^3+(aq) + e^- -> Fe^2+(aq) E standard = 0.770 V

Transcript

00:01 Hello everyone.

00:02 So for the given question, we have the standard reduction potential for each of the harsal reaction of overall cell reaction that is given over here.

00:10 The temperature is given as 221 calvines, but we need to determine is the equilibrium constant for this overall cell reaction.

00:19 Now first of all, from the given electrode potentials in the question, we can see that for the reaction 1, the electrode potential, that is the reduction potential, is higher, that is 1 .23 volts then for the second reaction that is 0 .770 volts and we know that higher the electrode reduction potential it means that higher is the tendency for that electrode to act as a cathode so first of all the first electrode that is the manganese dioxide will act as a cathode because it is having higher reduction potential and we know that at cathode the reduction process happens and the second the second electrode, that is for iron, will act as our anode and we know that at anode the oxidation reaction happens.

01:13 Now we will calculate the standard reduction potential for the cell.

01:18 So the standard reduction potential for the cell is given as the electrode potential for the cathode minus the electrode potential for the anode.

01:30 Now magnesium dioxide is acting as a cathode and for which the electrode potential is given as 1 .23 volts minus the electrode potential of anode will be 0 .770 volts.

01:42 So, electrode potential for the overall cell will be equal to 0 .46 volts.

01:48 Now from the given cell reaction, we can say that the overall number of moles of electron exchange will be equal to 2, that is, n in this case will be equal to 2.

02:01 And the relationship between the electrode potential of the cell and the equilibrium constant k is given as that e knot of the cell equals to rt divided by nf times natural log of the equilibrium constant k.

02:22 Now in this equation, this r is the gas constant and the value of gas constant to be taken is 8 .34 joules, kelvin inverse...

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