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Rufina [12.5K]
1 year ago
13

You are given the following three half-reactions:(1) Fe³⁺(aq) + e⁻ ⇄ Fe²⁺(aq) (2) Fe²⁺(aq) +2e⁻ ⇄ Fe(s) (3) Fe³⁺(aq) +3e⁻ ⇄ Fe(s

) (a) Use E°half-cell values for (1) and (2) to find E°half-cell for (3).
Chemistry
1 answer:
RideAnS [48]1 year ago
4 0

The E°half-cell of the (3) is 0.33 volts .

Given,

(1)  Fe3+(aq) +e =Fe2+(aq)

(2) Fe2+(aq) +2e = Fe(s)

(3) Fe3+(aq) +3e = Fe(s)

We know ,

E° half-cell of (1) is 0.77 volts .

E° half cell of ( 2) is -0.44 volts .

By resolving or adding equation 1 and 2 we will get (3) ,

Thus , the value of the E° half cell of (3) is given by ,

E°cell = E° half-cell of (1 ) + E° half-cell of (2 ) = 0.77-0.44 = 0.33 volts

Hence the E° half-cell of ( 3 ) is 0.33 volts .

<h3>What is cell potential? </h3>

It is the difference between the electrode potentials of two half cells.

it also defined as the force which causes the flow of electrons from one electrode to the another and thus results in the flow of current.

Ecell = E( cathode) - E ( anode)

<h3>What is electrode potential ? </h3>

The tendency of the electrode to lose of gain electrons is called as electrode potential.

Learn more about cell potential here :

brainly.com/question/19036092

#SPJ4

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An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
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Answer:

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Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

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W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

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Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of
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<u>Answer:</u> The smallest temperature change is shown by water.

<u>Explanation:</u>

To calculate the heat absorbed or released, we use the equation:

q=mC\times \Delta T      ......(1)

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q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

\Delta T = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

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We are given:

m=50.0g\\C_{Fe}=0.449J/g^oC

Putting values in equation 1, we get:

200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC

Change in temperature = 8.99°C

  • <u>Option b:</u>  50.0 g water, C_{water}=4.18J/g^oC

We are given:

m=50.0g\\C_{water}=4.18J/g^oC

Putting values in equation 1, we get:

200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC

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We are given:

m=50.0g\\C_{Pb}=0.128J/g^oC

Putting values in equation 1, we get:

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Change in temperature = 62.5°C

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We are given:

m=25.0g\\C_{Ag}=0.235J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC

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We are given:

m=25.0g\\C_{Fe}=0.79J/g^oC

Putting values in equation 1, we get:

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