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Ad libitum [116K]
3 years ago
7

The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate

ΔG for this reaction.
*** Please explain the reactions since I’m very confused as to wich side I should put the electrons.
Ex: Cu-> Cu2+ + 2e
Chemistry
1 answer:
Cloud [144]3 years ago
7 0

Answer : The \Delta G for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : 2Ag^++2e^-\rightarrow 2Ag

Now we have to calculate the Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o

where,

\Delta G^o = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole

Therefore, the \Delta G for this reaction is, -88780 J/mole.

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For the generic equilibrium HA(aq) ⇌ H+(aq) + A−(aq), which of these statements is true? For the generic equilibrium , which of
timama [110]

<u>Answer:</u> The correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

<u>Explanation:</u>

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of HA and KA follows the equation:

HA\rightleftharpoons H^{+}(aq.)+A^{-}(aq.)

KA\rightleftharpoons K^+(aq.)+A^{-}(aq.)

According to Le-Chateliers principle, if there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.

In the equilibrium reactions, A^- ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of HA.

Thus, the addition of KA will shift the equilibrium in the left direction.

Equilibrium constant depends on the temperature of the system. It does not have any effect on any change of pH.

pH is defined as the negative logarithm of hydrogen ions present in the solution

  • If the solution has high hydrogen ion concentration, then the pH will be low.
  • If the solution has low hydrogen ion concentration, then the pH will be high.

As, the equilibrium is shifting in the left direction, that means concentration of H^+ ions are getting decreases. This will increase the pH of the solution.

Hence, the correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

8 0
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The experimental apparatus represented above is used to demonstrate the rates at which gases diffuse. When the cotton balls are
cestrela7 [59]
The correct answer is —-C
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An analytical chemist is titrating 118.3 mL of a 0.3500 M solution of butanoic acid (HC3H7CO2) with a 0.400 M solution of KOH. T
TEA [102]

Answer:

pH = 12.33

Explanation:

Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.

The titration reaction is

HA + KOH ---------------------------- A⁻ + H₂O + K⁺

number of moles of HA :   118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA

number of  moles of OH  : 115.4 mL/1000ml/L x 0.400 mol/L  = 0.046 mol A⁻

therefore the weak acid will be completely consumed and what we have is  the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.

n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH

pOH = - log (KOH)

M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M

pOH = - log (0.0021) = 1.66

pH = 14 - 1.96 = 12.33

Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.

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What is the volume of the rock in the image in mL
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Answer:

Initial volume: mL

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Volume of rock: cm3

5 0
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