HCl gets oxidised

Oxidation involves increase in O.N while reduction involves decrease in O.N.

Let the oxidation number of 5 in H₂SO₄ be x
Write the oxidation number of each atom above its symbol
+1 x - 2
H₂SO₄
Calculate the sum of the oxidation numbers of all the atoms.
2 (+1) + x + 4 (-2) = 0
x - 6 = 0
x = +6
In Na₂SO₄
Write the oxidation number of each atom its symbol.
+1 x - 2
Na₂SO₄
Calculate the sum of the oxidation numbers of all the atoms.
2 (+1) + x + 4 (-2) = 0
2 + x-8 = 0
x = +6

A standard hydrogen electrode is called reversible electrode because it can react both as anode as well as cathode in an electrochemical cell

\(\overset { +1 }{ { H }_{ 3 } } \overset { x }{ P } { O }_{ 4 }^{ -2 }\) let the oxidation no. of P in H₃PO₄, be x.
Calculate the sum of the oxidation numbers of all the atoms
3 (+1) + x + 4 (-2) = 0
= 3 + x - 8 = x - 5 = 0
x = +5

Zero.

Oxidation involves loss of one or more electrons by a species during a reaction

A redox couple consists of oxidised and reduced form of the same substance taking part in the oxidation or reduction half reaction.

\(\overset { +1 }{ H2 } \overset { +2 }{ O } +\overset { O }{ { F }_{ 2 } } \longrightarrow \overset { +1 }{ H } \overset { -1 }{ F } +\overset { +1 }{ H } \overset { -2 }{ O } \overset { +1 }{ F } \)
Here, the O.N. of F decreases from 0 in F₂ to -1 in HF and increases from 0 in F₂ to +1 in HOF. Therefore, F₂ is both reduced as well as oxidised. Thus, it is a redox reaction and more specifically, it is a disproportionaiion reaction.

By conventional method. CH₃COOH = 2x + 4 - 4 = 0 or x = 0
By chemical bonding method, C2 is attached to three H-atoms (less electronegative than carbon) and one-COOH group (more electronegative than carbon).
\(H\overset { 2 }{ - } \overset { \overset { H }{ I } }{ \underset { \overset { I }{ H } }{ C } } -\overset { \overset { 0 }{ II } }{ C } -OH\)
therefore, O.N. of C₂ = 3 (+1) + x + 1 (-1) = 0 or x = -2
C1 is, however, attached to one oxygen atom by a double bond, one .O.H (O.N. = -1) and one CH₃ (O.N. = +1) group, therefore, O.N. of C₁ = + 1 + x + 1
(-2) + 1 (-1) = 0 or x = +2

Mn in K₂MnO₄
\(\overset { +1 }{ { K }_{ 2 } } \overset { x }{ Mn } \overset { -2 }{ { O }_{ 4 } } \)
2 (+1) + x + 4 (-2) = 0
x - 6 = 0
x = +6 oxygen.

Since, Mg²⁺(aq)/Mg electrode = - 2.36V is at lower potential than Al³⁺(aq)/AI electrode = - 1.66 V, therefore, Mg²⁺ (aq)/Mg electrode acts the anode and AI³⁺(aq)/Mg electrode acts as the anode and AI³⁺(aq) /AI acts as the cathode.In other words, Mg loses electrons and AI³⁺ ion accepts electrons.Thus, the cell reaction is
 3Mg +2 AI³⁺ \(\rightarrow\) 3Mg²⁺ + 2AI and  E^o_cell = E^o_{AL³⁺ I AI} - E_{Mg²⁺ I Mg} = -1.66-(-2.36)
= +0.70 V

Number of millimoles of K₂Cr₂O₇ present in 24 mL of 0.5 M solution = 24 x 0.5 =12.The balanced chemical equation for the redox reaction is
K₂Cr₂O₇ + 6(NH₄)₂ SO₄.6H₂O + 7H₂SO₄\(\rightarrow\)K₂SO₄ + 6(NH₄)₂SO₄ + 3Fe_2(SO4)₃ + Cr₂(SO₄)₃ + 43H₂O
From the balanved equation , 6 moles Mohr's salt are oxidised by 1 mole of K₂Cr₂O₇
12 millimoles of Mohr's salt will be oxidised by
= \(\frac { 1 }{ 6 } x12=millimoles \ K_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }\)

Since, E^o of Fe²⁺ /Fe (-0.44 V) is lower than that of Ag⁺ / Ag (+0.80 V) electrode, therefore, Ag⁺ gets reduced and Fe gets oxidised.As a result, concentration of Ag⁺ ions decreases while that of NO^-₃ ions remain unchanged.
 2Ag⁺ (aq) +Fe(s) \(\rightarrow\)  2Ag(s) + Fe²⁺ (aq)

 \(CI_{ 2 }O_{ 7 }(g)+4H_{ 2 }O_{ 2 }(aq)+2OH^{ - }(aq)\longrightarrow 2CIO_{ 2 }^{ - }(aq)+5H_{ 2 }O(I)+4O_{ 2 }(g)\)