Electrochemistry

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)

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.

The order in which the given metals displace each other from the solution of their salts is given by,
Mg>Al> Zn> Fe> Cu
A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt. The order of increasing the reducing power of given metals is Cu< Fe< Zn< Algiven metals displace each other from the solution of their salts is given by, Mg>Al> Zn> Fe> Cu. This is hence arranged in decreasing order of its reactivity

Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

Anode: Lead (Pb)

Cathode: a grid of lead packed with lead oxide (PbO₂)

Electrolyte: 38% solution of sulphuric acid (H₂SO₄)

The cell reactions are as follows :

Pb (s) + SO₂^-4 (aq) ⇒ PbSO₄ (s) + 2e^- (anode)

PbO₂ (s) + SO₂^-4 (aq) + 4H⁺ (aq) +2e^-⇒ PbSO₄ (s) +2H₂O (l) (cathode)

Pb (s) + PbO₂ (s) +2H₂SO₄ (aq)⇒ 2PbSO₄ (s) +2H₂O (l)

(overall cell reaction)

On charging, all these reactions will be reversed.

Cr₂O₇^2– + 14H⁺ + 6e^–⇒ 2Cr³⁺ + 7H₂O

A 3.12 

Cr₂O₇^2– + 14H⁺ + 6e^–⇒ 2Cr³⁺ + 7H₂O

For reducing one mole of Cr₂O₇^2–, 6 mole of electrons are required. Hence, 6 Faraday charges is needed. Hence, 6F = 6*96487 = 578922 C. Thus, the quantity of electricity is needed is 578922 C.

Metals with greater reactivity can be extracted electrolytically. Sodium, potassium, calcium, lithium, magnesium, aluminium which are present in the top of the reactivity series are extracted electrolytically. 

Current I = 0.5A
Time t = 2hrs = 2*60*60 = 7200 seconds
Charge Q = I * t
Q = 0.5*7200 = 3600 C
Charge carried by 1 mole of electrons (6.023*10²³electrons) is equal to 96487C.
No of electrons = 6.023*10²³ * 3600/96487
No of electrons = 2.25*10²² electrons

C = 0.025 mol L^-1
A_m = 46.1 Scm² mol L^-1
λ⁰ (H⁺) = 349.6 Scm² mol L^-1
λ⁰ (HCOO^-) = 54.6 Scm² mol L^-1
Λ⁰m (HCOOH) = Λ⁰ (H⁺) + Λ⁰ (HCOO^-) 
= 349.6 + 54.6
= 404.2 S cm² mol L^-1

^{Now, the degree of dissociation:}