Kelvi

Kelvi

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Kelvi

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Given below are half-cell reactions: $\\$ MnO $_4^- + 8H^+ + 5e^- \rightarrow$ Mn $^{2+} + 4H_2O, \quad E^\circ_{Mn^{2+}/MnO_4^-} = -1.510 V \\$ O $_2 + 2H^+ + 2e^- \rightarrow H_2O, \quad E^\circ_{0_2/H_2O} = +1.223 V \\$ Will the permanganate ion, MnO $_4^-$ liberate O $_2$ from water in the presence of an acid?

hard

Electrochemistry

2022

chemistry

Yes, because $E^\circ_{cell} = +0.287 V$

No, because $E^\circ_{cell} = -0.287 V$

Explanation

To determine if the permanganate ion MnO $_4^-$ can liberate O $_2$ from water, we need to calculate the $E^\circ_{cell}$

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Yes, because $E^\circ_{cell} = +2.733 V$

No, because $E^\circ_{cell} = -2.733 V$

for the overall reaction.

\\

The given half-cell reactions are:

\\

1.

MnO

_4^- + 8H^+ + 5e^- \rightarrow

Mn

^{2+} + 4H_2O, \quad E^\circ_{Mn^{2+}/MnO_4^-} = -1.510 \,

V

\\

2.

O

_2 + 2H^+ + 2e^- \rightarrow H_2O, \quad E^\circ_{O_2/H_2O} = +1.223 \,

V

\\

To find the overall cell potential

E^\circ_{cell},

we need to reverse the second reaction because we want

O

_2

to be produced.

\\

Reversed reaction:

\\

3.

2H_2O \rightarrow

O

_2 + 4H^+ + 4e^-, \quad E^\circ_{H_2O/O_2} = -1.223 \,

V

\\

Now, balance the number of electrons transferred in both reactions.

\\

Multiply the first reaction by 2 and the reversed second reaction by 5:

\\

4.

2(

MnO

_4^- + 8H^+ + 5e^- \rightarrow

Mn

^{2+} + 4H_2O) \\

5.

5(2H_2O \rightarrow

O

_2 + 4H^+ + 4e^-) \\

This gives:

\\

6.

2

MnO

_4^- + 16H^+ + 10e^- \rightarrow 2

Mn

^{2+} + 8H_2O \\

7.

10H_2O \rightarrow 5

O

_2 + 20H^+ + 20e^- \\

Combine the reactions:

\\

8.

2

MnO

_4^- + 16H^+ + 10H_2O \rightarrow 2

Mn

^{2+} + 8H_2O + 5

O

_2 + 20H^+ \\

Simplify the equation:

\\

9.

2

MnO

_4^- + 2H_2O \rightarrow 2

Mn

^{2+} + 5

O

_2 + 4H^+ \\

Calculate

E^\circ_{cell}: \\ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \\ E^\circ_{cell} = (-1.510 \,

V

) - (-1.223 \,

V

) \\ E^\circ_{cell} = -1.510 \,

V

+ 1.223 \,

V

\\ E^\circ_{cell} = -0.287 \,

V

\\

Since

E^\circ_{cell}

is negative, the reaction is not spontaneous.

\\

Therefore, the permanganate ion

MnO

_4^-

will not liberate

O

_2

from water.

\\

This corresponds to Option 2: No, because

E^\circ_{cell} = -0.287 \,

V

.