Class 12th

3.21 In the button cells widely used in watches and other devices the following reaction takes place: Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH– (aq) Determine ?r G ? and E ? for the reaction.

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Given - Zn → Zn²⁺ + 2e ^-, E⁰ = 0.76V (anode)

Ag₂O + H₂O + 2e ^- →2Ag + 2OH ^-, E⁰ = 0.344V (cathode), n = 2

Δ_rG⁰ =?

E_cell⁰=?

Zn is oxidized and Ag₂O is reduced.

Hence, the standard cell potential, E_cell⁰ is given as,

E_cell⁰ = ER⁰ - E_L⁰

E⁰ _cell = 0.344 + 0.76

∴ E⁰_cell = 1.104 V

To calculate the standard Gibb's free energy? _rG⁰, we use,

? _rG⁰ = - nE⁰F → Equation 1

= - 2*96487*1.104 J

= - 213043.296 J

∴? _rG⁰ = - 2.13*10⁵ J

The standard cell potential, E⁰_cell is 1.104 V and the standard Gibb's free energy? _rG⁰ is - 2.13*10⁵ J

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14.6 Why cannot vitamin C be stored in our body?

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14.6

Vitamin C (ascorbic acid due to extensive –H bonding with water due to the presence of –OH group) is a water-soluble vitamin in contrast to vitamin A, D, E and K which are fat soluble. Also, humans cannot synthesize it due to the lack of specific enzyme and it is rapidly absorbed from the intestine.

Because it is water-soluble, it is not stored in our body to a significant amount but is readily excreted in the urine.

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3.20 Write the Nernst equation and emf of the following cells at 298 K:

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A 3.5 Ecell = ?

(i) Mg + Cu²⁺ → Mg²⁺+ Cu (n = 2)

E⁰ Cu²⁺ / Cu⁺ = 0.34V

E⁰ Mg ²⁺ / Mg = - 2.37 V

E_cell⁰ = E_R⁰-E_L⁰

E_cell⁰ = 0.34 - ( - 2.37) → Equation 1

Using Nernst equation, we get,

Substituting Equation 1 in equation 2, we get,

∴ E_cell = 0.34 - ( - 2.37) - 0.0591 / 2 log 10 ^-3/ 10^-4

= 2.71 - 0.0591 / 2 log 10

= 2.71 - 0.02955

∴ E_cell = 2.68 V

The e.m.f of the cell, E_cell is 2.68 V

ii) Fe + 2H ⁺ →

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14.5 Where does the water present in the egg go after boiling the egg?

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14.5

Eggs contain protein and proteins are a polymer in amino acids. When an egg is boiled the protein present inside the egg is denatured (the process of loss of biological activity of protein like their ability of –H bonding when subjected to a physical change like a change in temperature, pH) and coagulated (the process of liquid changing to a solid or semi-solid state). Due to this, the water present in the egg is absorbed by the coagulated protein by –H bonding.

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3.19 Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd (ii) Fe 2+(aq) + Ag+ (aq) → Fe3+(aq) + Ag(s) Calculate the ?rG? and equilibrium constant of the reactions.

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(1) Known - E⁰_Cr³⁺_/Cr = - 0.74 V

E⁰ cd²⁺ = - 0.40 V

? _rG⁰ =? K =?

The galvanic cell of the given reaction is written as - Cr_(s)|Cr³⁺_(aq)| Cd²⁺_(aq)|Cd_(s)→ Reaction 1

Hence, the standard cell potential is given as, E⁰ = E_R⁰ - E_L⁰

= - 0.40 - (- 0.74)

∴ E⁰ = + 0.34 V

To calculate the standard Gibb's free energy? _rG⁰, we use,

? _rG⁰ = - nE⁰F → Equation 1

wherenF is the amount of charge passed and E⁰ is the standard reduction electrode potential. Substituting n = 6 (no. of e ^- involved in the reaction 1), F = 96487 C mol^-1,

E⁰ = + 0.34 V in Equation 1, we get, l

? _rG⁰ = - 6*0.34V*96487 C mol^-1

= - 196833.48 CV mol^-1

= - 196833.48 J mol^-1

∴?

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14.4 The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.

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14.4

Amino acids are organic compounds containing amine (basic) and carboxyl (acidic) functional group with a specific side chain. Both acidic and basic group are present in the same molecule. In, aqueous solution carboxyl group can lose a proton (H⁺) and amino group can accept a proton (H⁺) giving rise to the dipolar ion called as zwitter ion. Zwitter ion is shown below:

In this zwitter ion there is the presence of both positive as well as negative charge, so there is the development of strong electrostatic force of attraction between the molecules and the water. For this reason solubility of amino acids is higher. Due to strong e

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3.18 Depict the galvanic cell in which the reaction Zn(s)+2Ag+ (aq) →Zn2+(aq)+2Ag(s) takes place. Further show: (i) Which of the electrode is negatively charged? (ii) The carriers of the current in the cell. (iii) Individual reaction at each electrode.

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The galvanic cell corresponding to the given redox reaction can be represented as:

Zn|Zn²⁺_(aq)|Ag ⁺_(aq)|Ag

1) Zn electrode (anode) is negatively charged because, at this electrode, Zn is oxidized to Zn²⁺, causing electron accumulation at the
2) Electrons (ions) are the carriers of the current in the cell and in the external circuit, current flows from Ag (cathode) to Zn (anode) which is normally opposite to the electron flow which is from anode to cathode.
3) At anode:

Zn_(s)⇒ Zn2 + (aq) + 2e– At cathode:

Ag + (aq) + e –⇒ Ag (s)

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14.3 How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?

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14.3

D-glucose reacts with hydroxylamine (NH₂OH) to form oxime due to the presence of the aldehyde functional group (-CHO). This is due to the cyclic structure of glucose which forms an open chain structure in an aqueous medium, which then reacts to give an oxime.

But in case of pentaacetate of D-glucose, it does not form open chain structure in an aqueous medium so it does not react with NH₂OH.

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3.17 Given the standard electrode potentials, (given below) Arrange these metals in their increasing order of reducing power.

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K ⁺ /K = –2.93V, Ag⁺ /Ag = 0.80V, Hg²⁺/Hg = 0.79V Mg²⁺/Mg = –2.37 V, Cr³⁺/Cr = – 0.74V

A 3.2 Reducing power of metals increase with the decrease of reduction potential. Hence, the increasing order of reducing power will be as,
Ag < Hg < Cr < Mg < K
When the reduction potential is lower, the element has more tendency to get oxidized and thus more will be reducing power. The metal that has more negative electrode potential will be the one with more reducing power. Thus, here potassium (K) has the highest reducing power among the given elements.

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3.16 Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.

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