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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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3 0
3 years ago
Find the density of a material, given that a 5.01 g sample occupies 3.46 mL
Alex73 [517]

Answer:

The answer is

<h2>1.45 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}

From the question

mass of material = 5.01 g

volume = 3.46 mL

The density of the material is

density =  \frac{5.01}{3.46}  \\  = 1.44797687...

We have the final answer as

<h3>1.45 g/mL</h3>

Hope this helps you

3 0
3 years ago
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