No, the oxidation number of an element in any species is an apparent charge on the atom which it appears to have acquired when all other atoms in the species are removed as ions.

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
= \(\frac { 1 }{ 6 } x12=millimoles \ K_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }\)

Cu²⁺ do not oxidise benzaldehyde (C₆H₅CHO) to benzoic acid.
This includes that Ag⁺ is a stronger oxidising agent than Cu²⁺.

Pbo is a base. It reacts with nitric acid and forms soluble lead nitrate.
\(Pbo+ \ \underset { (acid-base \ reaction) }{ { 2HNO }_{ 3 } } \longrightarrow \ \underset { Soluble }{ Pb } \left( { NO }_{ 2 } \right) _{ 2 }+{ H }_{ 2 }O\)
Nitric acid does not reacts with Pbo₂ . Both of them are strong oxidising agents. In HNO₃ , nitrogen is in its maximum oxidation state(+5) and in Pbo₂ , lead is in its maximum oxidation sate(+4). Therefore, no reaction takes place.

O charge on the left, +6 charge on the right; add 6 electrons on the right side.

(a) MnO^-₄  loses oxygen, therfore it is an oxidant while C₂H₅OH gains oxygen, therefore it is a reductant.
(b) Zn is reductant whilr hydrogen (H) is oxidant
(c) pbO is oxidant while C is reductant

(a) In aqueous solution, AgNO₃ ionises to give Ag⁺(aq) and \(N{ O }_{ 3 }^{ - }\)(aq) ions.
AgNO₃(aq) \(\rightarrow\)  Ag⁺(aq) + \(N{ O }_{ 3 }^{ - }\)(aq)
Thus, when electricity is passed, Ag+(aq) ions move towards the cathode whie \(N{ O }_{ 3 }^{ - }\) ions move towards the anode. In other words, at the cathode, either Ag⁺(aq) ions or H₂O molecules may be reduced. Which of these will actually get discharged would depend upon their electrode potentials which are given below:
Ag⁺(aq) + e^- \(\rightarrow\) Ag(s); E^o = +0.80 V (i)
2H₂O(I) + 2e^- \(\rightarrow\) H₂(g) + 2OH^-(aq); EO = -0.83 V (ii)
Since the electrode potential (i.e., reduction potential of Ag+(aq) ions is higher than that of H₂O molecules, therefore, at the cathode, it is the Ag+(aq) ions (rather than H2O molecules) which are reduced.
Similarly, at the anode, either Ag metal of the anode or H₂O molecules may be oxidised. Their electrode potentials are:
Ag(s) \(\rightarrow\) Ag⁺(aq) + e^-; E^o = -0.80 V (iii)
2H₂O(l) \(\rightarrow\) O₂(g) + 4H⁺(aq) + 4e^-; E^o = -1.23 V (iv)
Since the oxidation potential of Ag is much higher than that of H₂O, therefore, at the anode, it is the Ag of the silver anode which gets oxidised and not the H₂O molecules. It may, however, be mentioned here that the oxidation potential of \(N{ O }_{ 3 }^{ - }\) ions is even lower than that of H2O since more bonds are to broken during reduction of \(N{ O }_{ 3 }^{ - }\) ions than those in H₂O. Thus, when an aqueous solution of AgNO₃ is electrolysed, Ag from Ag anode dissolves while Ag +(aq) ions present in the solution get reduced and get deposited on the cathode.
(b) If, however, electrolysis of AgNO₃ solution is carried out using platinum electrodes, instead of silver electrodes, oxidation of water occurs at the anode since Pt being a noble metal does not undergo oxidation easily. As a result, O₂ is liberated at the anode according to equation (iv). Thus, when an aqueous solution of AgNO₃ is electrolysed using platinum eectrodee, Ag+ ions from the solution get deposited on the cathode while o² is liberated at the anode.
(c) In aqueous solution, H₂SO₄ ionises to give H+(aq) and \({ SO }_{ 4 }^{ 2- }\)t(aq) ions.
H₂SO₄(aq) \(\rightarrow\) 7 2H⁺(aq) + \({ SO }_{ 4 }^{ 2- }\)(aq)
Thus, when electricity is passed, H+  (aq) ions move towards cathode while \({ SO }_{ 4 }^{ 2- }\) (aq) ions move towards anode. In other words, at cathode either H+(aq) ions or H₂O molecules are reduced. Their electrode potentials are:
2H⁺(aq) + 2e^- \(\rightarrow\)  H₂(g); E^o = 0.0 V
HzO(aq) + 2e- \(\rightarrow\) H₂(g) + 20H^-(aq); E^O = -0.83 V
Since the electron potential (i.e., reduction potential) of H+(aq) ions is higher than that of H₂O, therefore, at the cathode, it is W(aq) ions (rather than H₂O molecules) which are reduced to evolve H₂ gas.
Similarly at the anode, either \({ SO }_{ 4 }^{ 2- }\) (aq) ions or H₂O molecules are oxidised. Since the oxidation potential of \({ SO }_{ 4 }^{ 2- }\) is expected to be much lower (since it involved cleavage of many bonds as compared to those in H₂O than that of H₂O molecules, therefore, at the anode, it is H₂O molecules (rather than sot ions) which are oxidised to evolve O2 gas. From the above discussion, it follows that during electrolysis of an aqueous solution of H₂SO₄ only the electrolysis of H₂O occurs liberating H₂ at the cathode and O₂ at the anode.
(d) In aqueous solution, CuCl₂ ionises as follows:
CuCI₂(aq) \(\rightarrow\) Cu²⁺(aq) + 2C^- (aq)
On passing electricity, Cu²⁺(aq) ions move towards cathode and Cqaq) ions move towards anode.
Thus, at cathode, either Cu²⁺(aq) or H20 molecules are reduced. Their electrode
potentials are:
Cu²⁺(aq) + 2e^- \(\rightarrow\) Cu(s); E^o = +0.34 V
H₂O(I) + 2e^-  \(\rightarrow\) H₂(g) +2OH ; E^o = -0.83 V
Since the electrode potential of Cu ²⁺(aq)ions is much higher than that of H₂O,
therefore, at the cathode, it is Cu2+(aq) ions which are reduced and not H₂O molecules.
Similarly, at the anode, either C^-(aq) ions or H₂O molecules are oxidised. Their oxidation potentials are:
2C^-(aq) \(\rightarrow\)  Cl₂(g) + 2e^- \(\triangle\)E^o = -1.36 V
2H₂O(I) \(\rightarrow\) O₂(g) + 4H⁺(aq) + 4e^-; \(\triangle\)E^o = -1.23 V
Although oxidation potential of H₂O molecules is higher than that of Cl^- ions, nevertheless, oxidation of Cl^- ions occurs in preference to H₂O since due to overvoltage much lower potential than -1.36 V is needed for the oxidation of H₂O molecules.
Thus, when an aqueous solution of CuCl₂ is electrolysed, Cu metal is liberated at the cathode while Cl₂ gas is evolved at the anode.

\(4\overset { -3-1 }{ N{ H }_{ 3 } } (g)+5\overset { 0 }{ { O }_{ 2 } } (g)\longrightarrow 4\overset { +2-1 }{ NO } (g)+6\overset { +1-2 }{ H_{ 2 }O } (g)\)
Oxidation number of N increases from -2(in NH₃) to +2(in NO) and oxidation number of O decreases from zero (in 0₂) to-2
(in NO and H₂O). This shows that NH₃ is oxidised and O₂ is reduced. Hence, it is a redox reaction.

K\(\rightarrow \)K⁺e^- (oxidation half reaction)
\(\frac { 1 }{ 2 } \) Cl₂+e^-\(\rightarrow \)Cl^- (reduction half reaction)

Overall reaction: Ag(s) + Fe³⁺(aq) \(\rightarrow\) Ag⁺(aq) + Fe2+(aq); E^o= -0.03 V
Since the EMF of the reaction is negative, therefore, the above reaction is not feasible.
Alternatively, the reaction between Ag(s) and Fe³⁺(aq) may occur according to the following equation
3Ag(s) + Fe³⁺(aq) \(\rightarrow\) 3Ag⁺(aq) + Fe(s) 
On similar lines, we can calculate the e.m.f. of this reaction comes to be even more negative, i.e., -0.836 V, and hence this redox reaction is also not feasible.
(e) Suppose the reaction between Br₂(aq) and Fe²⁺(aq) occurs according to the following equation:
Br₂(aq) + 2Fe²⁺(aq) \(\rightarrow\) 2Br^-(aq) + 2Fe³⁺(aq)
The above reaction can be split into the following two half reactions. Writing electrode potential for each half reaction from the Table 8.1, we have
Oxidation: Fe₂₊(aq) \(\rightarrow\) Fe3+(aq) + e^-] x 2; E^o = -0.77 V
Reduction: Br₂(aq) + 2e^- \(\rightarrow\) 2Br-(aq); E^o = +1.09 V

At anode Zn\(\rightarrow \)Zn²+2e^-
At cathode \(\frac { Br_{ 2 }+2e^{ - }\rightarrow 2Br^{ - } }{ Zn+Br_{ 2 }\rightarrow Zn^{ 2 }+2Br^{ - } } \)