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Phoenix [80]
3 years ago
15

Be sure to answer all parts. Which product(s) form(s) at each electrode in the aqueous electrolysis of the following salts? Sele

ct as many as apply. (a) FeI2 Cathode: Anode: I− (aq) I2 (s) Fe (s) Fe2+ (aq) I− (aq) I2 (s) Fe (s) Fe2+ (aq) (b) K3PO4 Cathode: Anode: O2 (g) H2 (g) H+ (aq) OH− (aq) O2 (g) H2 (g) H+ (aq) OH− (aq)
Chemistry
1 answer:
zlopas [31]3 years ago
4 0

Answer:

FeI2

Anode

Fe^2+

Cathode

I2

K3PO4

Anode

H^+ and K^+

Cathode

H2

Explanation:

In electrolysis, there are two electrodes present, the cathode and the anode. The anode is the positive electrode. This is where oxidation occurs and positive ions are formed. Species give up a electrons at the anode and the electrons travel towards the cathode.

At the cathode (negative electrode where reduction occurs), species accept electrons according to their position in the electrochemical series. In the case of K3PO4, both hydrogen and potassium ions arrive at the cathode but hydrogen in preferentially discharged due to the position of the two ions in the electrochemical series.

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Answer:

0.1082M of Barium Hydroxide

Explanation:

KHP reacts with Ba(OH)2 as follows:

2KHP + Ba(OH)2 → 2H2O + Ba²⁺ + 2K⁺ + 2P²⁻

<em>Where 2 moles of KHP reacts per mole of barium hydroxide</em>

<em />

To solve this question we must find the moles of KHP in 1.37g. With these moles and the reaction we can find the moles of Ba(OH)2 and its molarity using the volume of the solution (31.0mL = 0.0310L) as follows:

<em>Moles KHP -Molar mass: 204.22g/mol-</em>

1.37g * (1mol / 204.22g) = 0.006708 moles KHP

<em>Moles Ba(OH)2:</em>

0.006708 moles KHP * (1mol Ba(OH)2 / 2mol KHP) =

0.003354 moles Ba(OH)2

<em>Molarity:</em>

0.003354 moles Ba(OH)2 / 0.0310L =

<h3>0.1082M of Barium Hydroxide</h3>
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