Given the following half-reactions, predict the species that will be obtained at

Question

Given the following half-reactions, predict the species that will be obtained at the cathode and anode for the electrolysis of an aqueous solution of potassium iodide KI(aq):

O2(g) + 4 H+ (aq) + 4 e – 2 H2O() E o = +1.229 V

I2(s) + 2 e – 2 I (aq) E o = +0.535 V

2 H2O() +2 e H2(g) + 2 OH (aq) E o = 0.828 V

K + (aq) + e K(s) E o = 2.925 V

(a) O2(g) at the cathode and H2O() at the anode.

(b) H2(g) at the cathode and O2(g) at the anode.

(c) K(s) at the cathode and O2(g) at the anode.

(d) K(s) at the cathode and I2(s) at the anode.

(e) H2(g) at the cathode and I2(s) at the anode.

Explanation / Answer

O2(g) + 4 H+ (aq) + 4 e – 2 H2O() E o = +1.229 V
I2(s) + 2 e – 2 I (aq) E o = +0.535 V
2 H2O() +2 e H2(g) + 2 OH (aq) E o = 0.828 V
K + (aq) + e K(s) E o = 2.925 V

The reaction with maximum Eo occurs at cathode
O2(g) + 4 H+ (aq) + 4 e – 2 H2O()
H2O will accepet e- and H2 will be released at cathode

At anode:
K(s) K + (aq) + e
At anode K will be consumed

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