All categories

2 years ago

What is the overall cell potential for this redox reaction?

1 answer:

3 0

Based on the given chemical reaction, Barium undergoes oxidation since it losses electrons while Ca undergoes reduction since it gains electrons. On the given single REDOX reaction, barium gains electrons and undergo reduction, this has to be reversed to satisfy the net reaction, therefore the cell potential becomes +2.90. By this the over-all cell potential = 0.03 V

You might be interested in

Someone please help me with this

Inessa05 [86]

Answer:

With what???

Explanation:

I don't see anything.

5 0

2 years ago

I WILL MARK BRAINIEST PLEASE HELP QUICKLY!!

omeli [17]

Answer:

6CO2 + 6H2O → C6H12O6 + 6O2 is a balanced equation

Explanation:

Carbon, Oxygen, Hydrogen

Reactants are carbon dioxide and water, products are glucose and oxygen

Dynamic Equilibrium I have no idea about (soz)

I believe this might be endothermic

4 0

2 years ago

A liquid has_____volume and a _____ shape.

makvit [3.9K]

A liquid has definite volume but no definite shape .

5 0

2 years ago

For the reaction CO2(g) + H2(g)CO(g) + H20(g)

Studentka2010 [4]

Answer:

The ΔG° is 29 kJ and the reaction is favored towards reactant.

Explanation:

Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,

ΔG° = ΔH°rxn - TΔS°rxn

= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K

= 41.2 kJ - 12.2 kJ

= 29 kJ

As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.

5 0

2 years ago

In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char

lisabon 2012 [21]

Answer : The partial pressure of $N_2,H_2\text{ and }NH_3$ at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

$N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)$

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of $K_p$ will be,

$K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$

Now put all the values of partial pressure, we get

$0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}$

By solving the term x, we get

$x=0.287\text{ and }3.889$

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of $NH_3$ at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of $N_2$ at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of $H_2$ at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of $N_2$ + Partial pressure of $H_2$ + Partial pressure of $NH_3$

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0

3 years ago