Animal and Plant Cells Worksheet Word

A voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn

nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_{[Ni^{2+}/Ni]}=-0.23V$

$E^0_{[Zn^{2+}/Zn]}=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^{2+}+2e^-$ $E^0_{[Zn^{2+}/Zn]}=-0.76V$

Reaction at cathode (reduction) : $Ni^{2+}+2e^-\rightarrow Ni$ $E^0_{[Ni^{2+}/Ni]}=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}$

$E^o=(-0.23V)-(-0.76V)=0.53V$

(a) Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}$

$E_{cell}=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M.

New concentration of $Ni^{2+}$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}$

$E_{cell}=0.52V$

(c) Now we have to calculate the concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}$

Now put all the given values in the above equation, we get:

$0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}$

$x=1.49M$

The concentration of $Ni^{2+}$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

5 0

3 years ago

The density of helium in a 2.00 L tank at 1.0 atm and 23 degrees Celsius is?

andrezito [222]

The output density is given as kg/m 3, lb/ft 3, lb/gal(US liq) and sl/ft 3. Specific weight is given as N/m 3 and lb f / ft 3.

5 0

3 years ago

Air is made up mostly of O2, N2, and CO2. If the total pressure of air is 765mmHg and P(N2) = 0.79 atm and P(CO2) = 0.05 atm, th

Mrac [35]

Answer:

the partial pressure of the O2 is 0.167 atm

Explanation:

The computation of the partial pressure of the O2 is shown below:

As we know that

P = P_N2 + P_O2 + P_CO2

P_O2 = P - P_N2 - P_CO2

= (1.007 - 0.79 - 0.05)

= 0.167 atm

Hence, the partial pressure of the O2 is 0.167 atm

we simply applied the above formula

5 0

3 years ago

A gas has a volume of 27.5 L at 302 K and 1.40 atm. How many moles are in the sample of gas?

ioda

Answer:

1.55 mol

Step-by-step explanation:

We can use the Ideal Gas Law to solve this problem.

pV = nRT Divide both sides by RT

n = (pV)/(RT)

Data:

p = 1.40 atm

V = 27.5 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 302 K

Calculations:

n = (1.40 × 27.5)/(0.082 06 × 302)

= 1.55 mol

6 0

3 years ago

The specific heats and densities of several materials are given below: Material Specific Heat (cal/g·°C) Density (g/cm3) Brick 0

abruzzese [7]

Answer: The change in temperature is 84.7°C

Explanation:

To calculate the change in temperature, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed = 1 kCal = 1000 Cal (Conversion factor: 1 kCal = 1000 Cal)

m = mass of steel = 100 g

c = specific heat capacity of steel = 0.118 Cal/g.°C

$\Delta T$ = change in temperature = ?

Putting values in above equation, we get:

$1000cal=100g\times 0.118cal/g^oC\times \Delta T\\\\\Delta T=\frac{1000cal}{100g\times 0.118cal/g^oC}\\\\\Delta T=84.7^oC$

Hence, the change in temperature is 84.7°C

5 0

3 years ago

Animal and Plant Cells Worksheet Word ​

Animal and Plant Cells Worksheet Word