Electrochemistry Test - 4

Based on electro chemical series, oxidising power of
F₂ > Cl₂ > Br₂ > I₂

On passing Cl₂ in to a solution containing KF, Kl and KBr,
2KBr + Cl₂ → 2KCI + Br₂ (orange)
2KI + Cl₂ → 2KCI + l₂ (violet)
2KI + Br₂ → 2KBr + l₂ (violet)
Br₂ formed also oxidises Kl to l₂ (violet)

E° _cell = 1.10 V
(a) Hence , when E_ext < 1.10 V, reaction continues till the E_ext reaches 1.10 V and
Zn(s) + Cu²⁺(aq) → Cu(s) + Zn²⁺(aq)
thus , zinc dissolves at anode and Cu deposits at cathode.
(b) When E_ext = 1.10 V , no reaction.
(c) When E_ext > 1.10V, reverse reaction takes place,

It now acts as electrolytic cell.
Zinc deposits at zinc plate which is now the cathode (+ ve plate) and Cu dissolves as Cu²⁺ at copper plate which is now the anode (- ve plate)

Based on electro chemical series, reactivity series is

(a) 2Ag⁺ + Cu → Cu²⁺ 2Ag
Cu is a better reducing agent than Ag hence Ag⁺ is reduced by Cu. Thus AgNO₃ can not be stored in copper vessel.
(b) Cu²⁺ + Mg → Mg²⁺ + Cu
Mg will reduce Cu²⁺ to Cu thus can not be stored.
(c) Cu²⁺ + 2Ag → Cu + 2 Ag⁺
Cu is better reducing agent hence CuCI₂ can be stored in silver vessel.
(d) Hg²⁺ + Cu → Cu²⁺ + Hg
Cu is better reducing agent. Hence, HgCI₂ can not be stored in copper vessel. 

Oxides of the metals are decomposed to metals if in electro chemical series , E^o_red > 0 .7 9 V

For A g, E°_Ag⁺/Ag = 0.80 V hence Ag₂O is decomposed to metals. Products are Ag , CO₂ and O₂. Since reaction takes place in an open vessel, CO₂ and O₂ escape from mixture, residue being 2 moles of silver.

Blue colour fades or changes to colourless if Cu²⁺ is reduced to Cu.

In electrochemical series (ECS), reactivity order is Zn, Fe, Cu, Ag

Thus, Zn and Fe reduce Cu²⁺ to Cu but Ag does not reduce Cu²⁺. Thus, in I and III, CuSO₄ changes to colourless or fades by zinc and iron plates.

3Mn²⁺(aq) + 2MnO₄^-(aq)+ 2H₂O (/) → 5MnO₂(s) + 4H⁺  E°_cell = 0.47 V
Since E°_cell > 0, hence spontaneous.
Thus, Mn²⁺ is oxidised to MnO₂ by MnO₄^- in acidic medium.

E°_cell  remains unchanged on division , thus intensive property.

Δ_rG is dependent on the division , thus extensive property.

Since E° _cell < 0, disproportionation of Sn²⁺ to Sn⁴⁺ (by oxidation) and Sn (by reduction) is not spontaneous.

It is oxidation half-cell

H₂ (reduced part) is at 1 bar. H⁺ (oxidised part) is at 1 M. 

It is reduction half-cell

In this case also, oxidised and reduced parts are at unity.

Then (I), (II) and (III) are correct facts.

In ECS, pair with more negative values of E°red reducing agent is above oxidising agent.
Thus, Z/Z^- is the best reducing agent

Thus, Z^- will reduce both X and Y and itself will be oxidised to Z .

E°_x = - 0.52 V
E°_y = -3.03 V
E°_z = - 1.18 V

Their placements in ECS is in order Y > Z > X.
Thus, reducing nature is also in same order y > Z > X

These metals based on E°_red values are placed in the following order:

Mg will reduce and Ag⁺ to metals Cu, Hg, Ag.
Cu will reduce and Ag⁺ to Hg, Ag, Hg will reduce Ag⁺ to Ag.
Ag⁺ is reduced by all metals thus Ag first. Hg²⁺ is reduced by Cu, Mg thus Hg. Cu²⁺ is reduced by Mg only, thus Cu.
Thus, deposition : 

In electrochemical series (ECS), elements have been arranged in the increasing standard reduction potential starting from most negative to most positive value w.r.t. SHE (E°_she= 0.00 V).

Most negative standard reduction potential means most easily oxidised and thus is the best reducing agent.

OR Reducing agent in (ECS) is always above oxidising agent. Thus, Zn(s) is the best reducing agent in given set.

Mn³⁺ + e^- → Mn²⁺ + 1.57 V
Fe³⁺ + e^- → Fe²⁺ + 0.77 V
Co³⁺ + e^- → Co²⁺ + 1.97 

Cr³⁺/Cr²⁺ with most negative E°_red is the best reducing agent.
Cr²⁺→ Cr³⁺ + e^- , E° = 0.41 V
Thus, is oxidised from Cr²⁺ to Cr³⁺ most easily.

Most negative E°_red means the Mⁿ⁺/M is at the top of ECS and is the best reducing agent.

E°_Zn²⁺/Zn = -0.76 V
E° _Cu²⁺/Cu = + 0.34 V

In electrochemical series, zinc is above copper and thus zinc is a better reducing agent than copper. When CuS0₄ is placed in zinc vessel, copper is displaced

Thus, CuSO₄ cannot be stored in a vessel made of zinc.
Thus, statement I is incorrect and statement II is correct.

In electrochemical series, copper is above silver thus is a better reducing agent. When AgNO₃ solution is stirred with copper spoon, Ag is displaced and copper is oxidised to Cu²⁺ (blue).

Thus, statement I and II are correct and statement II is the correct explanation of statement I.

E°_cell > 0 thus spontaneous
Cu²⁺/Cu couple is thus a stronger oxidising agent. Reverse reaction is non-spontaneous.
Cu + 2H⁺ → Cu²⁺ + H₂, E°_cell = - 0.34 V
Thus, copper cannot displace H₂ from acid.

At cathode reduction occurs, therefore, cathode acts as a reducing agent by supplying electrons.

Hence, option B is correct.

According to Faraday's laws, for deposition of 1 gm of an equivalent weight of a substance, amount of charge required is equal to the charge on one mole of electrons.

i.e., the Electric charge required = Electric charge on one mole of electrons

Thus, (i) → (p)
(ii) (B) O₂(g) + 2H₂O + 4e^- → 40H^- n₁ = 4, E₁ = 0.401 V
(C) O₂(g)+4H⁺ 4e^- → 2H₂O , n₂ = 4, E₂ = 1.229 V
(G) 4H⁺ + 4e^- → 2H₂ , n₃ = 4, E₃ = 0.00 V
(B) + (G) - (C) gives
4 H₂O + 4e^- → 2H₂ + 40H^- 
2 H₂O + 2e^- → H₂ + 2OH^- 
∴ E₄ = 0.401 + 0.00- 1.229
= -0.83 V
Thus, (ii) → (s)




Thus, (iv) → (s)

For (I) and (II), E° > 0, but for reduced species.
For (III), E° > 0, thus Cr³⁺ is the most stable.
More negative value of standard electrode reduction potential, better the reducing agent. Thus,
Mn²⁺ < CI^- < Cr³⁺ < Cr

For (I) and (II), E° > 0, but for reduced species.
For (III), E° > 0, thus Cr³⁺ is the most stable.
More negative value of standard electrode reduction potential, better the reducing agent. Thus,
Mn²⁺ < CI^- < Cr³⁺ < Cr

CH₃CH₂OH is oxidised to CH₃COOH and is reduced to Cr³⁺.
E°_cell - E°CH₃CH₂OH/CH₃COOH + E°/Cr³⁺
= -0.06+1.33 
= 1.27 V
Reaction quotient

CH₃CH₂OH is oxidised to CH₃COOH and is reduced to Cr³⁺.
E°_cell - E°CH₃CH₂OH/CH₃COOH + E°/Cr³⁺
= -0.06+1.33 
= 1.27 V
Reaction quotient

The apparatus shown is divided into two types of cells separated by slate walls. The first type, shown on the right and left of the diagram, uses an electrolyte of sodium chloride solution, a graphite anode (A), and a mercury cathode (M). The other type of cell, shown in the center of the diagram, uses an electrolyte of sodium hydroxide solution, a mercury cathode (M), and an iron anode (D).