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Nina [5.8K]
2 years ago
13

Consider the reaction in the lead-acid cell pb(s) pbo2(s) 2 h2so4(aq) 2 pbso4(aq) 2 h2o(l) for which e°cell = 2. 04 v at 298 k.

δg° for this reaction is:________
Chemistry
1 answer:
aleksley [76]2 years ago
8 0

δg° for the reaction having e° cell = 2.04v. is 393.658kJ.

The lead (Pb) in the lead storage battery act as an anode

<h3>What is Voltaic Cell?</h3>

A voltaic cell or galvanic cell is an electrochemical cell that uses chemical reactions to produce electrical energy.

<h3>Parts of voltaic cell</h3>

Anode where oxidation occurs Cathode where reduction occurs.

Therefore, Pb acts as anode and PbO2 act as a cathode.

Given,

E° cell = 2.04v

Faraday constant= 96485 C/mol

Moles of electron transferred n= 2

∆G = -nFE°

= -2 × 96485 × 2.04

= 393.658kJ

Thus we concluded that the δg° for the reaction having e° cell = 2.04v. is 393.658kJ.

learn more about voltaic cell :

brainly.com/question/1370699

#SPJ4

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3 0
3 years ago
Can somebody please help me with this question, I'm very confused! I thought i did it right because i found Kb and the conc. of
andrew-mc [135]
Let's investigate the substances involved in the reaction first. The compound <span>CH3NH3+Cl- is a salt from the weak base CH3NH2 and the strong acid HCl. When this salt is hydrated with water, it will dissociate into CH3NH2Cl and H3O+:

CH3NH3+Cl- + H2O </span>⇒ CH3NH2Cl + H3O+

Nest, let's apply the ICE(Initial-Change-Equilibrium) table where x is denoted as the number of moles used up in the reaction:

                 CH3NH3+Cl- + H2O ⇒ CH3NH2Cl + H3O+
Initial                  0.51                             0               0
Change                 -x                             +x             +x
-------------------------------------------------------------------------------
Equilibrium        0.51 - x                         x               x

Then, let's find the equilibrium constant of the reaction. Since the reaction is hydrolysis we use KH, which is the ratio of Kw to Ka or Kb. Kw is the equilibrium constant for water hydrolysis which is equal to 1×10⁻¹⁴. Since the salt comes from the weak base, we use Kb. Since pKb = 3.44, then. 3.44 = -log(Kb). Thus, Kb = 3.6307×10⁻⁴ 

KH = Kw/Kb = (x)(x)/(0.51 - x)
1×10⁻¹⁴/ 3.6307×10⁻⁴ = x²/(0.51-x)
x = 3.748×10⁻⁶

Since x from the ICE table is equal to the equilibrium concentration of H+, we can find the pH of the aqueous solution:

pH = -log(H+) = -log(x)
pH = -log ( 3.748×10⁻⁶)
pH = 5.43
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3 years ago
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