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Taya2010 [7]
2 years ago
5

A constant current of 0.350 A is passed through an electrolytic cell containing molten CrCl₂ for 21.7 h. What mass of Cr(s) is p

roduced? The molar mass of chromium is 52.0 g/mol. Provide your answer rounded to 3 significant digits.​
Chemistry
1 answer:
zhuklara [117]2 years ago
4 0

An electrochemical cell can generate or use electrical energy. The mass of solid chromium that will be deposited on the electrochemical plate is 7.17 gm.

<h3>What is current?</h3>

Current in an electrochemical cell is the ratio of the quantity of electricity in columns and time in seconds.

Given,

Current (I) = 0.350 A

Time = 21.7 hours

Molar mass of chromium = 52.0 g/mol

First time is converted into seconds:

1 hour = 3600 seconds

21.7 hours = 76020 seconds

The quantity of electricity flowing in the electrochemical solution is calculated as:

\begin{aligned} \rm Q & = \rm It\\\\& = 0.350 \times 76020 \\\\& = 26607\;\rm C \end{aligned}

Electricity required for depositing 1 mole or 52.0 g chromium is calculated as:

In electrochemical solution, chromium chloride is dissociated as:

\rm CrCl_{2} \rightarrow Cr^{2+} + 2 Cl^{-} \\\\\rm Cr^{2+} +2 e^{-} \rightarrow Cr

Two moles of electrons are needed to deposit 52.0 g of chromium.

If, 1 electron = 96500 C

Then, 2 electron = 193000 C

The mass of chromium deposited is calculated as:

193000 C = 52 g chromium

So, 26607 C = \dfrac{26607 \times 52}{193000} = 7.17 \;\rm gm

Therefore, 7.17 gm of chromium is produced.

Learn more about an electrochemical cell here:

brainly.com/question/20355190

#SPJ1

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w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

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