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1. Assign oxidation number to the underlined elements in each of the following species
(a) NaH2PO4
(b) NaHSO4
(c) H4_P2O7
(d) K2MnO4
(e) CaO2
(f) NaBH4
(g) H2S2O7
(h) KAl(SO4)2.12H2O
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Solution:
(a) NaH2PO4 = 1(+1) + 2(+1) + P + 4(-2) = 0
1+2+P-8=0
P-5=0
∴ O. N. of P = +5
(b) NaHSO4 = 1(1) + 1(1) + S + 4(-2) = 0
2+S-8=0
∴ O. N. of S = +6
(c) H4 P2O7 = 4(H) + 2(P) + 7(O) = 0
4(1) + 2P + 7(-2) = 0
4 + 2P - 14 = 0
2P - 10 = 0
∴ O. N. of P = +5
(d) K2 Mn O4 = 2(k) + Mn + 4(0) = 0
2(1) + Mn + 4(-2) = 0
2 + Mn - 8 = 0
∴ O. N. of Mn = +6
.in
(h) K Al(SO4)2 . 12 H2O
1(K) + 1(Al) + 2(S) + 8(O) = 0
1(1) + 1(3) + 2(S) + 8(-2) = 0
4 + 2S - 16 = 0
2S = +12
∴ O. N. of S = +6.
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(g) H2S2O7 = 2(H) + 2(S) + 7(0) = 0
2(1) + 2S + 7(-2) = 0
2 + 2S - 14 = 0
2S = +12
∴ O. N. of S = +6
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(f) Na = 1 ; H = -1 ( it’s a hydride)
1 × 1 + B€× 1 + 4€×€-1 = 0
1+B-4=0
B=3
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(e) Ca02 = 1(Ca) + 2(0) = 0
1(2)+2(0) = 0
2(0) = -2
∴ O. N. of O = -1
2. What are the oxidation numbers of the underlined elements in each of the following
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and how do you rationalize your results?
(a) KI3 (b) H2_S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH
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Solution:
(a) KI3 = 1(K) + 3(I) = 0
1(1) + 3I = 0
3I = -1
∴ O. N. of I = -1/3
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(b) H2S4O6 = 2(H) + 4(S) + 6(O) = 0
2(1) + 4S + 6(-2) = 0
4S - 10 = 0
S=
=
∴ O. N. of S = +5/2
(c) Fe3O4 = 3Fe + 4(0) = 0
3Fe + 4(-2) = 0
3Fe = +8
Fe=+8/3
∴€O. N. of Fe=+8/3
(d) CH3CH2OH = C2H6O
2C + 6(H) + 1(O) = 0
2C + 6(1) + 1(-2) = 0
2C + 4 = 0
C = -2
.in
Solution:
(a) and (b) on the basis of classical idea.
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3. Justify that the following reactions are redox reactions
(a) CuO(s) + H2(g)
Cu(s) + H2O(g)
(b) Fe2O3(s) + 3CO(g)
2Fe(s) + 3CO2(g)
(c) 4BCl3(g) + 3LiAlH4(s)
2B2H6(g) +3LiCl(s) + 3 AlCl3(s)
(d) 2K(s) + F2(g)
2K+F-(s)
(e) 4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
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(e) CH3COOH = C2H4O2
2C + 4(H) + 2(0) = 0
2C + 4(1) + 2(-2) = 0
2C + 4 - 4 = 0
2C = 0
C=0
∴ O. N. of C = 0
The above values are calculated on the basis of conventional methods.
(c ) on the basis of new concept of electrodensity.
(d ) on the basis of change in O.N.
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(e) Both classical as well as change of electron density.
4. Fluorine reacts with ice and results in the change H2O(s) + F2(g)
HOF(g) Justify that this reaction is a redox reaction.
HF(g) +
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Solution:
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We can explain this on the basis of change in oxidation no. F in HF and HOF. The oxidation no. of
fluorine changes from 0 to -1 in HF (reduced) whereas it increases from 0 to +1 in
HOF(oxidized).Hence this reaction is redox as well as a disproportionation reaction.
5. Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O72and NO3-. Suggest structure of these compounds. Count for the fallacy.
Solution:
H2SO5 = 2(H) + 1(S) + 5(O) = 0
2(1) + S + 5(-2) = 0
2 + S - 10 = 0
∴ S = +8
∴€O. N. of S = +8
.in
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6. Write formulas for the following compounds
(a) Mercury (II) Chloride (b) Nickel (II) sulphate
(c) Tin (IV) oxide (d) Thallium (I) sulphate
(e) Iron (III) sulphate (f) Chromium (III) oxide.
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NO-3 = 1(N) + 3(0) = -1
N + 3(-2) = -1
N - 6 = -1
N = -1 + 6
∴€O. N. of N = +5.
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Cr2O7 2- = 2(Cr) + 7(0) = -2
2Cr + 7(-2) = -2
2Cr = -2 + 14 = 12
∴ O. N. of Cr = +6
Solution:
(a) HgCl2
(b) NiSO4
(c) SnO2
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(d) Tl2SO4
(e) Fe2(SO4)3
(f) Cr2O3
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7. Suggest a list of the substances where carbon can exhibit oxidation states from –4 to
+4 and nitrogen from –3 to +5.
Solution:
Carbon can exhibit oxidation states from -4 to +4 in the following compounds.
The compounds of nitrogen with oxidation state from –3 to 5 are
NH3, NH2-NH2, NH=NH, N N, N2O, NO, N2O3, N2O4, N2O5 are the compounds of nitrogen where the
oxidation state of nitrogen is -3,-2,-1,0,1,2,3,4,5 respectively.
(i) sulphur in SO2 has an oxidation state of +4. The maximum oxidation state of S is +6 (e.q., SO3)
and minimum is –2(H2S). Similarly, H2O2 has oxygen atom in oxidation state of oxygen is +2 and
minimum oxidation state is –2 (in oxides, H2O, etc.)
(ii) In HNO3 nitrogen is in its maximum oxidation state of +5 and hence can only act as oxidant.
Similarly oxidation state of oxygen in ozone is zero and hence can change to its stable oxidation
state of –2 i.e., act as an oxidant.
8. While sulphur dioxide and hydrogen per oxide can act as oxidising as well as reducing
agents in their reactions, ozone and nitric acid act only as oxidants. Why?
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Solution:
Any substance can act as an oxidising and reducing agent if one of the element in the compound is
present in an intermediate oxidation state so that it can either increase its oxidation number
(reducing agent) or decrease its oxidation number (oxidising agent).
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sulphur in SO2 has an oxidation state of +4. The maximum oxidation state of S is +6 (e.q., SO3) and
minimum is –2(H2S). Similarly, H2O2 has oxygen atom in oxidation state of oxygen is +2 and
minimum oxidation state is –2 (in oxides, H2O, etc.)
.in
In HNO3 nitrogen is in its maximum oxidation state of +5 and hence can only act as oxidant.
Similarly oxidation state of oxygen in ozone is zero and hence can change to its stable oxidation
state of –2 i.e., acts as an oxidant.
9. Consider the reactions
a. 6CO2(g) + 6H2O(l)
C6H12O6(aq)+ 6O2(g)
b. O3(g) + H2O2(l) H2O(l) + 2O2(g) Why it more appropriate to write these reactions as:
a)6CO2(g) + 12H2O(l)
C6H12O6(aq)+ 6H2O(l) )+ 6O2(g)
b)O3(g) + H2O2(l) H2O(l) + O2(g) + O2(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Solution:
4
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a) 6CO2(g) + 12H2O(l)
C6H12O6(aq)+ 6H2O(l) )+ 6O2(g)
The above reaction is an overall photosynthetic reaction. It is appropriate to write it this way because
12 H2O are used per molecule of carbohydrate formed and 6 H2O are produced as the product of
photosynthesis.
b) O3(g)
O2(g)+[O]
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Hence it is more appropriate to write the reaction as 2 O2 molecules are produced by two different
reactants.
O3(g) + H2O2(l) H2O(l) + O2(g) + O2(g)
10. The compound AgF2 is an unstable compound. However, if formed, the compound
acts as a very strong oxidising agent. Why?
Solution:
AgF2 can be synthesized by fluorinating Ag2O with elemental Flourine.
AgF2 is light sensitive and reacts vigorously with water and hence it is considered as an unstable
compound. It should be stored in Teflon, a passivated metal container, or a quartz tube.
AgF2 acts as a very strong oxidising agent because Ag is in +2 oxidation state in AgF2 which is very
unusual. Ag in AgF2 is a powerful electron (1e-) acceptor to get converted to Ag1+ state hence
making AgF2 a powerful oxidising agent.
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11. How do you account for the following observations?
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are
used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic
potassium permanganate as an oxidant. Why? Write a balanced redox equation for the
reaction.
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(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride,
we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we
get red vapours of bromine. Why?
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Solution:
(a) Alcoholic KMnO4 is used as an oxidant in the manufacture of benzoic acid because organic
reactions must be performed in a medium of organic solvent (like alcohol) in which the oxidizing
agent (inorganic salt) is able to dissolve.
(b) When conc.H2SO4 is added to an inorganic mixture containing chloride, a red vapour of bromine
and SO2(colourless) is evolved. HBr can be formed only if dil.H2SO4 is used.
12. Identify the substances oxidized and reduced, the oxidising agent and
reducing agent for each of the following reactions:
(a) 2AgBr (s) + C6H6O2(aq)
2Ag(s) + 2HBr (aq)+ C6H4O2(aq)
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(b) HCHO(l) + 2[Ag (NH3)2]+ (aq) + 3OH- (aq)
2Ag(s) + HCOO-(aq) + 4NH3(aq) +
2H2O(l)
(c) HCHO (l) + 2Cu2+ (aq) + 5 OH-(aq)
Cu2O(s) + HCOO-(aq) + 3H2O(l)
(d) N2H4(l) + 2H2O2(l)
N2(g) + 4H2O(l)
(e) Pb(s) + PbO2(s) + 2H2SO4(aq)
2PbSO4(s) + 2H2O(l)
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Solution:
Substances which undergo oxidation act as a good reducing agent while the ones which undergoes
reduction act as good oxidising agent.
(a) Silver is reduced while C of compound C6H6O2 is oxidised. Therefore, oxidising agent is Ag and
reducing agent is C6H6O2.
(b) HCHO undergoes oxidation while [Ag (NH3)2]+(Tollens reagent) undergoes reduction.
Oxidising agent is Tollen’s reagent; reducing agent is HCHO.
(c) HCHO undergoes oxidation while Cu2+ (from Fehling’s solution) undergoes reduction.
∴€oxidizing agent is Fehling’s solution and reducing agent is HCHO.
(d) Nitrogen in N2H4 undergoes oxidation (reducing agent), H2O2.undergoes reduction (oxidising
agent).
(e) Pb undergoes oxidation (red. Agent) ; PbO2 undergoes reduction (oxidising agent).
13. Consider the reactions
2S2O32-(aq) + I2(s) → S4O62-(aq) + 2I- (aq)
S2O32-(aq) + 2Br2(l) + 5H2O(l) → SO42-(aq) + 4Br-(aq) + 10H+(aq)
Why does the same reductant, thiosulphate react differently with iodine and bromine?
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Solution:
Bromine acts as an oxidizing agent. It oxidizes sodium thio sulphate to sodium sulphate.
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S2O32-(aq) + 2Br2(l) + 5H2O(l) → SO42-(aq) + 4Br-(aq) + 10H+(aq)
2Na2S2O3(aq) + I2(s) → Na2S4O6(aq) + 2NaI (aq)
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Iodine being a weaker oxidizing agent reacts with thio sulphate and forms sodium tetrathionate.
.in
14. Justify giving reactions that among halogens, fluorine is the best oxidant and among
hydrohalic compounds, hydroiodic acid is the best reductant.
Solution:
Fluorine is the best oxidant among all halogens due to its smaller atomic size and greater nuclear
attraction it attracts electrons more readily than other halogens.
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Hydroiodic acid is the best reductant.
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Due to the greater size of I2 Hydrogen can be easily knocked out from HI hence it will be a supplier
of H+ ion and therefore used as reducing agent.
Because Cl2 is more electronegative than iodine.
15. Consider the reactions
(a) H3PO2(aq) + 4 AgNO3(aq) + 2H2O(l) → H3PO4(aq) + 4Ag(s) +4HNO3(aq)
(b) H3PO2(aq) + 4 CuSO4(aq) + 2H2O(l) → H3PO4(aq) + 2Cu(s) +H2SO4(aq)
(c) C6H5CHO(l) + 2[Ag(NH3)2]+(aq) + 3OH-(aq) → C6H5COO-(aq) + 2Ag(s)+ 4NH3(aq) +
2H2O
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(d) C6H5CHO (l) + 2Cu2+(aq) + 5OH-(aq) → No change observed
What inference do you draw about the behaviour of Ag+ and Cu2+ from these reactions?
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Solution:
In reactions (a) and (c). Ag+ is acting as electron acceptor ie, oxidising agent and get converted to
Ag.
In reaction c [Ag(NH3)2]+ (aq) is reduced to metallic silver.
In reaction (d) Cu2+ remains unchanged i.e, neither reduced nor oxidized.
.in
In reaction (b) Cu2+ is converted to Cu ie, 2e-s are accepted by Cu2+ion. Therefore it played the role
of oxidising agent.
16. Balance the following redox reactions by ion- electron method
(a) MnO4-(aq) + I- (aq)
MnO2 (s) + I2(s) (in basic medium)
(b) MnO4- (aq) + SO2 (g)
Mn2+ (aq) + HSO4- (aq) (in acidic solution)
2+
(c) H2O2 (aq) + Fe (aq)
Fe3+(aq) + H2O (l) (in acidic solution)
(d) Cr2O72- + SO2(g)
Cr3+(aq) + SO42-(aq) (in acidic solution)
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Solution:
(a) 2MnO4- + 6I- + 4H2O
2MnO2 + 3I2 + 8OH(b) 5SO2 +2MnO4 + 2H2O + H+
5HSO4- + 2Mn+2
+2
+
+3
(c) 2Fe + H2O2 + 2H
2Fe +2H2O
(d) 3SO2 + Cr2O72- + 2H+
2Cr+3 + 3SO42- +H2O
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17. Balance the following equations in basic medium by ion- electron method and
oxidation number methods and identify the oxidising agent and the reducing agent.
(a) P4(s) +OH-(aq)
PH3(g) + HPO2- (aq)
(b) N2H4(l) + ClO3 (aq)
NO(g) + Cl- (g)
(c) Cl2O7(g) + H2O2(aq)
ClO2-(aq) + O2(g) + H+
Solution:
(a) 4P + 3OH- + 3H2O
PH3 + 3H2PO2(b) 3N2H4 + 4ClO3
4Cl + 6H2O + 6NO
(c) Cl2O7 + 4H2O2 + 2OH2ClO2- + 5H2O + 4O2
18. What sorts of informations can you draw from the following reaction?
(CN)2(g) + 2OH-(aq)
CN-(aq) + CNO-(aq) + H2O(l)
Solution:
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Informations
19. Consider the elements:Cs, Ne, I and F
.in
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1. One cyanide molecule in (CN)2 is undergoing reduction by accepting one electron forming CN- ion.
2. Another Cyanide molecule in (CN)2 is undergoing oxidation by losing one electron forming CNOion.
3. The oxygen atom in CNO- ion is supplied by OH- ion.
4. The H2O molecules formed is due to the presence of OH-ions on the reactant side.
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only positive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor does the positive oxidation
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state.
Solution:
(a) I can exhibit –ve oxidation state as it is highly electronegative element.
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(b) Cs is highly electropositive element and exhibits +ve oxidation state.
(c) I can show both -ve as well as +ve oxidation states. For example, in HI it is –1 and Icl3 it is +3.
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(d) Ne is a noble gas with stable electronic configuration and hence neither exhibits –Ve nor +ve
oxidation state.
20. Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of
chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this
redox change taking place in water.
Solution:
The balanced equation is
SO2 + H2O → H2SO3
H2SO3 + NaOH →€NaHSO3 + H2O
(or)
HSO3 + NaOH → Na2SO3 + H2O
Na2SO3 + Cl2 +H2O → Na2SO4 + 2HCl.
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21. Refer to the periodic table given in your book and now answer the following
questions:
(a) Select the possible non metals that can show disproportionation reaction.
2H2O2 → 2H2O + O2
Cu2O + H2SO4 → Cu + CuSO4 + H2O
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Cl2 + 2NaOH → NaClO + NaCl +H2O
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Solution:
(a) 3NaOH + P4 + H2O → 3NaH2PO2 + PH3
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(b) Select three metals that can show disproportionation reaction.
3S + NaOH + H2O → Na2SO3 + 2H2S
3NO2 + H2O → 2H+ + 2NO3- +NO
Disproportionation reaction is a redox reaction in which some atoms of a single element in a reactant
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are oxidized and others are reduced.
(b) The above mentioned reactions are a few examples of disProportionation reactions. Cu, Mn,
Sn are the metals which can undergo dispoportionation reaction because they do exhibit variable
oxidation states.
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Cl, O, I, S, N, P are some of the possible non-metals which can undergo disproportionation reaction.
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22. In Ostwald’s process for the manufacture of nitric acid, the first step involves the
oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the
maximum weight of nitric oxide that can be obtained starting only with 10.00g. of ammonia
and 20.00g of oxygen?
Solution:
4NH3 + 5O2
4NO + 6H2O
Molecular mass of NH3 = 17g mol-1
Molecular mass of O2 = 32g mol-1
Molecular mass of NO = 30g mol-1
(4
17 =) 68g of NH3 reacted with (5 × 32 =) 160g of O2 to form (4 × 30=)120g of NO.
68 g of NH3 reacts with oxygen = 160 g
∴10 g of NH3 react with O2 =
= 23.5 g
According to the equation, 160 g of O2 produce NO = 120 g
= 15 g.
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∴20 g of O2 will produce No =
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The quantity of O2 available is less than what is required for the reaction.. Hence O2 is the limiting
reagent and the calculations must be based on the quantity of O2.
Solution:
Zn
+ Cu2+
Reducing
Oxidising
Zn2+ + Cu
.in
23. 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.
Agent
agent
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2Al
+ 3Zn2+
Reducing
agent
2Al3+ + 3Zn
Oxidising
Agent
w
decreasing order of reducing strength.
09
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
Al > Zn> Cu
24. Given the standard electrode Potentials, K+/K = -2.93V, Ag+/Ag = 0.80V,
Hg2+/Hg = 0.79V
Mg2+/Mg = -2.37V. Cr3+/Cr = -0.74V
Arrange these metals in their increasing order of reducing power.
Solution:
The increasing order of reducing power is given below
Ag+ Ag = 0.8 V < Hg2+/Hg = 0.79v < Cr3+/Cr = - 0.74V < Mg2+/Mg = -2.37
V<K+/K = -2.93.
25. Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq)
takes place, further show
Zn2+ (aq)+2Ag(s)
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell, and
(iii) Individual reaction at each electrode.
Solution:
(i) Ag electrode is negatively charged
2Ag+ +2e-
2Ag
Ag
.in
on
At cathode, Ag+
Zn2+ + 2e-
ti
(iii) At anode, Zn
ca
(ii) Migration of Ag+ions towards cathode
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DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES
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