Ask question

Ask question

All categories

1 month ago

16

The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when

[ Zn2+] = 3.5 M and [Pb2+] = 2.0⋅10−4 M.
Pb2+ (aq) + Zn (s) → Zn2+ (aq) + Pb (s)
A) 0.84
B) 0.76
C) 0.50
D) 0.63
E) 0.39

1 answer:

castortr0y [3K]1 month ago

5 0

Answer: The cell potential for the given reaction stands at 0.50 V

Explanation:

The provided cell reaction is:

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

The half-reactions are:

Anode oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Cathode reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

Initially, we need to find the cell potential for this reaction.

Utilizing the Nernst equation:

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}$

where,

F = Faraday's constant = 96500 C

R = gas constant = 8.314 J/mol·K

T = room temperature = $25^oC=273+25=298K$

n = electrons exchanged in oxidation-reduction = 2

$E^o_{cell}$ = standard electrode potential for the cell = +0.63 V

$E_{cell}$ = cell potential for the reaction =?

$[Zn^{2+}]$ = concentration of Zn²⁺ = 3.5 M

$[Pb^{2+}]$ = $2.0\times 10^{-4}M$

Now substituting all known values into the equation, we arrive at:

$E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}$

$E_{cell}=0.50V$

So, the resulting cell potential for this reaction is 0.50 V

You might be interested in

A 250 ml flask contains 3.4 g of neon gas at 45°c. Calculate the pressure of the neon gas inside the flask.

eduard [2782]

The solution to your inquiry yields P = 17.73 atm. Explanation: The volume V is 250 ml, equivalent to 0.25 liters (L), with a mass of 3.4 g and a temperature of 45°C, which converts to 318°K. We utilize the ideal gas law PV = nRT for the calculations.

7 0

1 month ago

The ions Ca2+ and PO43- form salt with the formula

Alekssandra [3086]

Ca3(PO4)2 is the correct formula.

7 0

2 months ago

Read 2 more answers

What mass of water in grams will fill a tank 100 cm long, 50 cm wide, and 30 cm high? Knowing that the density of water is 1 g/m

castortr0y [3046]

The mass is 150,000 grams. Multiply 100 by 50 by 30 to determine the container's volume, which equals 150,000 cm^3. Since a milliliter is equivalent to one cubic centimeter, and given that the density of water is one gram per milliliter, it follows that the mass of water is 150,000 grams.

7 0

1 month ago

A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

KiRa [2933]

Response:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$

Clarification:

Weight of the alloy $m_{a}$ = 25 gm

Initial temperature $T_{a}$ = 100°c = 373 K

Weight of the water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 25.32 °c = 298.32 K

Final temperature $T_{f}$ = 27.18 °c = 300.18 K

Using the energy balance equation,

Heat released by the alloy = Heat absorbed by the water

$m_{a}$ $C_{a}$ [[ $T_{a}$ - $T_{f}$ ] = $m_{w}$ $C_w$ ( $T_{f}$ - $T_{w}$ )

25 × $C_{a}$ × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

$C_{a} = 0.37 \frac{KJ}{Kg K}$

This gives us the specific heat of the alloy.

4 0

2 months ago

27. How much energy is required to change 150.0 g of water from 10.0°C to 45.0°C? (Cwater =

Alekssandra [3086]

Answer:

The correct option is C. 21900.3. I calculated 21945 J, which makes option C closely aligned with my result.

Explanation:

Data

mass = 150 g

initial temperature T1 = 10°C

final temperature T2 = 45°C

Cw = 4.18 J/g°C

Formula

Q = mCΔT = mC(T2 - T1)

Substitution

Q = (150)(4.18)(45 - 10)

Simplification

Q = (150)(4.18)(35)

Result

Q = 21945 J

5 0

1 month ago