Changes in the forms of energy are called

In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined

faltersainse [42]

Answer:

For A: The standard cell potential of the reaction is 4.4 V

For B: The standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C: The reaction is spontaneous as written.

Explanation:

For A:

The given chemical reaction follows:

$2Li(s)+Cl_2(g)\rightarrow 2Li^+(aq.)+2Cl^-(aq.)$

The given half reaction follows:

Oxidation half reaction: $Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V$ ( × 2)

Reduction half reaction: $Cl_2(g)+2e^-\rightarrow 2Cl^-(aq.);E^o_{Cl_2/2Cl^-}=+1.36V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

Here, chlorine will undergo reduction reaction will get reduced.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

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

$E^o_{cell}=1.36-(-3.04)=4.4V$

Hence, the standard cell potential of the reaction is 4.4 V

For B:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

n = number of electrons transferred = $2mol\text{ e}^-$

F = Faradays constant = $96500J/V.mol\text{ e}^-$

$E^o_{cell}$ = standard cell potential = 4.4 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J$

Hence, the standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C:

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.

7 0

3 years ago

How to balance the chemical equation: CO2+H2O-->C6H12O6

Andru [333]

Answer:

6H₂O + 6CO₂ + energy → C₆H₁₂O₆ + 6O₂

Explanation:

The given reaction represent the formation of glucose so it is photosynthesis reaction.

Photosynthesis:

It is the process in which in the presence of sun light and chlorophyll by using carbon dioxide and water plants produce the oxygen and glucose.

Carbon dioxide + water + energy → glucose + oxygen

water is supplied through the roots, carbon dioxide collected through stomata and sun light is capture by chloroplast.

Chemical equation:

6H₂O + 6CO₂ + energy → C₆H₁₂O₆ + 6O₂

it is known from balanced chemical equation that 6 moles of carbon dioxide react with the six moles of water and created one mole of glucose and six mole of oxygen.

4 0

3 years ago

What is the element that is in period 5 and group 3?

rjkz [21]

Answer:

Rubidium

Rubidium is the first element placed in period 5.

6 0

3 years ago

A 25.0-mL sample of 0.150 M butanoic acid is titrated with a 0.150 M NaOH solution. What is the pH before any base is added? The

hoa [83]

HA ⇄ H⁺ + A⁻
so:
$\frac{[H^+][A^-]}{[HA]} = 1.5 x 10^{-5}$
and now:
$\frac{(x)(x)}{(0.150-x)}$ = 1.5 x 10⁻⁵
x is considered very small compared to 0.15
x² = 2.25 x 10⁻⁶
x = 1.5 x 10⁻³
So [H⁺] = 1.5 x 10⁻³
pH = - log [H⁺] = - log (1.5 x 10⁻³) = 2.83

7 0

3 years ago

Which of the following is not a correct chemical equation for a double displacement reaction?

Step2247 [10]

Answer:

B. CaCl + LiCO3 yields CaCO3 + LiCl is not correct

It should be CaCl2 + Li2CO3 → 2LiCl + CaCO3

Explanation:

For a reaction to be double displacement reaction there are two things we need to look for

1) There must be an interchange of the group of ions

2) The reactants must dissolve in water to release ions

A. 2RbNO3 + BeF2 yields Be(NO3)2 + 2RbF

2Rb+ + NO3- + Be^2+ + 2F- → Be(NO₃)₂ + 2RbF

This is correct

B. CaCl + LiCO3 yields CaCO3 + LiCl

This is not correct

The correct equation is:

CaCl2 + Li2CO3 → Ca2+ + 2Cl- + 2Li+ + CO3^2- → 2LiCl + CaCO3

C. Na3PO4 + 3KOH yields 3NaOH + K3PO4

3Na+ + PO4^3- + 3K+ + 3OH- → 3NaOH + K3PO4

This is correct

D. 2MgI2 + Mn(SO3)2 yields 2MgSO3 + MnI4

2Mg^2+ + 4I- + Mn^4+ + 2SO3^2- → 2 MgSO3 + MnI4

This is correct

8 0

3 years ago