A positively charged sodium ion and a negatively charged chlorine ion are joined by

Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh

Dima020 [189]

Answer: The $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

Explanation:

For the given chemical reaction:

$2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)$

Half reactions for the given cell follows:

Oxidation half reaction: $ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V$ ( × 2)

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

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

$E^o_{cell}=1.36-(0.954)=0.406V$

To calculate standard Gibbs free energy, we use the equation:

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

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = 0.406 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J$

Hence, the $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

4 0

3 years ago

hi, can you check if my answers are correct? I'm not so confident in this topic (topic: Classifying Types of Chemical Reactions)

NikAS [45]

I am pretty sure your answer are correct , from what I know. Good job!

7 0

3 years ago

PLEASE HELP!!

wariber [46]

Answer:

answer is option 2 hope it helps

8 0

2 years ago

One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 C, and the temperature af

Licemer1 [7]

Answer:

fH = - 3,255.7 kJ/mol

Explanation:

Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

4 0

3 years ago

What is the suffix for - ✯ Alcohols ✯ Ketones ✯ Halogens thankyou ~

antoniya [11.8K]

Answer:

ane, al, keto

ol, al, keto

ol, al, one

ol, ane, one.

Explanation:

The suffix –ol is used in organic chemistry principally to form names of organic compounds containing the hydroxyl (–OH) group, mainly alcohols (also phenol). The suffix was extracted from the word alcohol. The suffix also appears in some trivial names with reference to oils (from Latin oleum, oil).

Functional group is a ketone, therefore suffix = -one

Hydrocarbon structure is an alkane therefore -ane

The longest continuous chain is C5 therefore root = pent

The first point of difference rule requires numbering from the left as drawn to make the ketone group locant 2-

pentan-2-one or 2-pentanone

CH3CH2CH2C(=O)CH3

7 0

2 years ago

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