Download App

Electro Chemistry — Chemistry STD 12 — Question

Tamilnadu BoardEnglish MediumSTD 12ChemistryElectro Chemistry5 Marks
Question
Derive the relationship between Gibb’s free energy and maximum work obtained from galvanic cell and equilibrium constant.
✓

Answer

1. In a galvanic cell, chemical energy is converted into electrical energy. The electrical energy produced by the cell is equal to the product of the total charge of the electrons and emf of the cell which drives these electrons between the electrodes.
2. If ' $n$ ' is the number of moles of electrons exchanged between the oxidising and reducing agent in the overall cell reaction, then the electrical energy produced by the cell is given as below.
Electrical energy $=$ change of ' $n$ ' moles of electrons $x E_{\text {cell }}$
Charge of $I$ mole of electrons $=$ one Faraday $=1 F$
Charge of $n$ moles of electrons $= nF$
3. Electrical energy $= nFE _{\text {cell }}$
This energy is used to do electric work. Therefore the maximum work that can be obtained from a galvanic cell is
Here the - ve sign is introduced to indicate that work is done by the system on the surroundings.
4. Second law of thermodynamics states that the maximum work done by the system is equal to the change in Gibbs free energy of the system
$
W _{\max }=\Delta G \text {. }
$
5. $\Delta G =- nFE _{\text {cell }}(6)$
For a spontaneous cell reactions, the AG should be negative. The above expression indicates that $E _{ cell }$ should be positive to get a negative AG value.
6. When

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "[ 5 Marks Questions ]" from Electro Chemistry→Electro Chemistry — all question types→Chemistry STD 12 question papers→Tamilnadu Board STD 12 question papers→Tamilnadu Board question paper generator→

Similar questions

Complete the reaction.
(i) $H _3 BO _3 \stackrel{373 K }{\longrightarrow} A + H _2 O$
(ii) $A \stackrel{413 K }{\longrightarrow} B + H _2 O$
(iii) $B \stackrel{\text { Red hot }}{\longrightarrow} C + H _2 O$
→
An unknown aldehyde 'A' on reacting with alkali gives a b - hydroxy - aldehyde, which loses water to form an unsaturated aldehyde, 2 - butenal. Another aldehyde 'B' undergoes disporportionation reaction in the presence of conc. alkali to form products C and D. C is - an aryl alcohol with formula $C _7 H _8 O$.
1. Identify A and B .
2. Write the sequence of reactions involved
3. Name the product, when 'B' reacts with Zn amalgam and hydrochloric acid.
→
Explain about the cyclic structure of Glucose.
→
Discuss about the hydrolysis of salt of weak acid and weak base and derive pH value for the solution.
→
Write short note on metal excess and metal deficiency defect with an example.
→
Write a notes on
  1. Sheet silicates
  2. Three dimensional silicate
→
Explain the mechanism of aldol condensation.
→
Explain about the hydrolysis of salt of strong acid and weak base. Derive $K _{ h }$ and pH for that solution.
→
Derive an expression for the Nernst equation.
→
The rate law for a reaction of $A, B$ and $C$ has been found to be rate $=k[A]^2[B][L]^{\frac{3}{2}}$. How would the rate of reaction change when
i. Concentration of $[ L ]$ is quadrupled
ii. Concentration of both $[A]$ and $[B]$ are doubled
iii. Concentration of $[A]$ is halved
iv. Concentration of $[ A ]$ is reduced to $\left(\frac{1}{3}\right)$ and concentration of $[ L ]$ is quadrupled.
→