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GalinKa [24]
2 years ago
13

A concentration cell is constructed by using the same half-reaction for both the cathode and anode. What is the value of standar

d cell potential, for a concentration cell that combines a silver anode in contact with 0.10 M silver nitrate and a silver cathode in contact with 0.00003 M silver nitrate
Chemistry
1 answer:
Akimi4 [234]2 years ago
7 0

Solution :

A cell that is concentrated is constructed by the same half reaction for the anode as well as he cathode.

We know,

In a standard cell,

the reduction half cell reaction is :

$Ag^+(aq)+e^- \rightarrow Ag(s) E^0 = -0.80 \ V$

The oxidation half ell reaction :

$Ag(s) \rightarrow Ag^+(aq) + e^- \ E^0= +0.80 \ V$

Thus the complete reaction of the cell is :

$Ag^+(aq)+ Ag(s) \rightarrow Ag^+(aq)+Ag(s)$

$E^0 $ cell = $E_R - E_L = 0.00  \ \text{volts}$

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Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions. If 0.212 g of KHP are dissolve
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0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.

Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions.

The balanced neutralization equation is:

NaOH(aq) + KHC₈H₄O₄(aq) ⇒ KNaC₈H₄O₄(aq) + H₂O(l)

  • Step 1: Calculate the reacting moles of KHP.

0.212 g of KHP react. The molar mass of KHP is 204.22 g/mol.

0.212 g × 1 mol/204.22 g = 1.04 × 10⁻³ mol

  • Step 2: Determine the reacting moles of NaOH.

The molar ratio of NaOH to KHP is 1:1.

1.04 × 10⁻³ mol KHP × 1 mol NaOH/1 mol KHP = 1.04 × 10⁻³ mol NaOH

  • Step 3: Calculate the molarity of NaOH.

1.04 × 10⁻³ moles of NaOH are in 35.00 mL of solution.

[NaOH] = 1.04 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.0297 M

0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.

Learn more about titration here: brainly.com/question/4225093

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