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3 years ago

10

Electrolysis is used in the electroplating of metals. The same amount of current is passed through separate aqueous solutions of

CuSO4, Sn(SO4)2 and Cr2(SO4)3 in separate electrolytic cells for the same amount of time. State and explain which cell would deposit the greatest amount (in mol) of metal. Identify the electrode at which the metal is deposited.

1 answer:

Aleonysh [2.5K]3 years ago

3 0

Answer:

CuSO4 cell will have the greatest amount of deposit among all three. The deposit will occur at the cathode

Explanation:

The valence of the elements in this case is as follows -

Cu - 2e-

Sn - 4e-

Cr - 3e-

CuSO4 cell will have the greatest amount of deposit among all three

The atoms of copper metal will deposit at the cathode. At the cathode, the least number of moles of electrons needed .

Hence, more amount of copper can be extracted out by the electrolyte

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Which equation correctly relates the heat of reaction to the standard heats of formation?

VARVARA [1.3K]

Answer:

D

Explanation:

∆H° = ∆Hf ° (products) – ∆Hf ° (reactants)

4 0

3 years ago

The fact that HBO2, a reactive compound, was produced rather than the relatively inert B2O3 was a factor in the discontinuation

Reil [10]

Answer:

1027.62 g

Explanation:

For $B_2H_6$ :-

Mass of $B_2H_6$ = 296.1 g

Molar mass of $B_2H_6$ = 27.66 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{296.1\ g}{27.66\ g/mol}$

$Moles\ of\ B_2H_6= 10.705\ mol$

From the balanced reaction:-

$B_2H_6(g) + 3 O2_{(l)}\rightarrow 2 HBO_2_{(g)}+ 2 H_2O_{(l)}$

1 mole of $B_2H_6$ react with 3 moles of oxygen

Thus,

10.705 mole of $B_2H_6$ react with 3*10.705 moles of oxygen

Moles of oxygen = 32.115 moles

Molar mass of oxygen gas = 31.998 g/mol

<u>Mass = Moles * Molar mass = 32.115 * 31.998 g = 1027.62 g</u>

8 0

3 years ago

This substance consists of two or more different elements chemically combined

kow [346]

The substance would be referred to as a compound substance.

8 0

3 years ago

Calculate the pOH of a solution if the concentration of hydroxide ions (OH-) is 1.9 x 10-5M?

OLEGan [10]

Answer:

9.28

Explanation:

pOH refers to a measure of hydroxide ions concentration. pOH tells about the alkalinity of a solution. If pOH is less than 7 then aqueous solutions are alkaline, acidic if pOH is greater than 7 and neutral if pOH is equal to 7.

Concentration of the hydroxide ions = 1.9 x 10-5 M

pH = $-log(1.9\times 10^{-5})=4.72$

pOH = 14 - pH

=14 - 4.72 = 9.28

6 0

4 years ago

A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th

Mashcka [7]

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$ .

(a) Relation between $K_{a}$ and $pK_{a}$ is as follows.

$pK_{a} = -log (K_{a})$

Putting the values into the above formula as follows.

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Also, relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log\frac{[conjugate base]}{[acid]}$

= $3.347+ log \frac{0.15}{0.12}$

= 3.44

Therefore, pH of the buffer is 3.44.

(b) No. of moles of HCl added = $Molarity \times volume$

= $11.6 M \times 0.001 L$

= 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form $HNO_{2}$

Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

= 0.15 mol

And, no. of moles of $HNO_{2}$ = $0.12 M \times 1.0 L$

= 0.12 mol

On the other hand, after the reaction :

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

= (0.15 - 0.0116) mol

= 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

= (0.12 + 0.0116) mol

= 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = $pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

= 3.369

= 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0

3 years ago