Q. No.The values of x and y in the following redox reaction are:
\(\mathrm{xCl}_2+6 \mathrm{OH}^{-} \rightarrow \mathrm{ClO}_3^{-}+\mathrm{yCl}^{-}+3 \mathrm{H}_2 \mathrm{O}\)
1. x = 2 , y = 4
2. x = 5 , y = 3
3. x = 3 , y = 5
4. x = 4 , y = 2
\(\mathrm{xCl}_2+6 \mathrm{OH}^{-} \rightarrow \mathrm{ClO}_3^{-}+\mathrm{yCl}^{-}+3 \mathrm{H}_2 \mathrm{O}\)
1. x = 2 , y = 4
2. x = 5 , y = 3
3. x = 3 , y = 5
4. x = 4 , y = 2
Subtopic: Balancing of Equations |
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Level 1: 80%+
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For the given redox reaction, find out the coefficients a, b and c of a balanced equation respectively:
\(\mathrm{{a} {Cr}_2 {O}_7^{2-}+{bSO}_3^{2-}({aq})+{c} {H}^{+}({aq})} →\)\(\mathrm{2 {a} {Cr}^{3+}({aq})+{bSO}_4^{2-}({aq})+\frac{{c}}{2} {H}_2 {O}(\ell)} \)
\(\mathrm{{a} {Cr}_2 {O}_7^{2-}+{bSO}_3^{2-}({aq})+{c} {H}^{+}({aq})} →\)\(\mathrm{2 {a} {Cr}^{3+}({aq})+{bSO}_4^{2-}({aq})+\frac{{c}}{2} {H}_2 {O}(\ell)} \)
| 1. | 8, 1 and 3 | 2. | 1, 3 and 8 |
| 3. | 3, 8 and 1 | 4. | 1, 8 and 3 |
Subtopic: Balancing of Equations |
78%
Level 2: 60%+
NEET - 2023
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In the following balanced reaction,
\(\small{\begin{aligned} &\mathrm{XMnO}_4^{-}+\mathrm{YC}_2 \mathrm{O}_4^{2-}+\mathrm{ZH}^{+}\rightleftharpoons \mathrm{XMn}^{2+}+2 \mathrm{YCO}_2+\frac{\mathrm{Z}}{2} \mathrm{H}_2 \mathrm{O} \end{aligned}}\)
values of \(X\), \(Y\) and \(Z\) respectively, are:
1. 2, 5, 16
2. 8, 2, 5
3. 5, 2, 16
4. 5, 8, 4
\(\small{\begin{aligned} &\mathrm{XMnO}_4^{-}+\mathrm{YC}_2 \mathrm{O}_4^{2-}+\mathrm{ZH}^{+}\rightleftharpoons \mathrm{XMn}^{2+}+2 \mathrm{YCO}_2+\frac{\mathrm{Z}}{2} \mathrm{H}_2 \mathrm{O} \end{aligned}}\)
values of \(X\), \(Y\) and \(Z\) respectively, are:
1. 2, 5, 16
2. 8, 2, 5
3. 5, 2, 16
4. 5, 8, 4
Subtopic: Balancing of Equations |
78%
Level 2: 60%+
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In the following equation, the products A and B are, respectively are:
\(\small Cr_2O^{2-}_7 (aq) + 3SO^{2-}_3 (aq) + 8H^+(aq) \rightarrow A + B + 4H_2O(l)\)
1. \(2 \mathrm{Cr}^{2+} ; 3 \mathrm{SO}_2 \)
2. \(2 \mathrm{Cr}^{+} ; \mathrm{S}_2 \mathrm{O}_7{ }^{2-} \)
3. \(2 \mathrm{Cr}^{3+} ; 3 \mathrm{SO}_4{ }^{2-} \)
4. \(2 \mathrm{Cr}^{3+} ; 3 \mathrm{HSO}_4{ }^{-}\)
Subtopic: Balancing of Equations |
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Level 1: 80%+
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Consider the following reaction:
Pb3O4 + 4HNO3 → A + B + 2H2O
A and B are respectively-
1. A=Pb(NO3)4; B=PbO2
2. A=Pb; B=O2
3. A= Pb(NO3)2; B=PbO2
4. A=PbO; B=NO2
Pb3O4 + 4HNO3 → A + B + 2H2O
A and B are respectively-
1. A=Pb(NO3)4; B=PbO2
2. A=Pb; B=O2
3. A= Pb(NO3)2; B=PbO2
4. A=PbO; B=NO2
Subtopic: Oxidizing & Reducing Agents | Balancing of Equations |
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Level 2: 60%+
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In a reaction, \(BrO^-_3 \) changes into \(Br_2 \) in the acidic medium. The number of electrons that are needed to balance the equation are:
1. 10 electrons on the left-hand side
2. 6 electrons on the left-hand side
3. 3 electrons on the left-hand side
4. 3 electrons on the right-hand side
1. 10 electrons on the left-hand side
2. 6 electrons on the left-hand side
3. 3 electrons on the left-hand side
4. 3 electrons on the right-hand side
Subtopic: Balancing of Equations |
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Level 2: 60%+
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What is the coefficient for H+ when the half equation is balanced with the smallest whole number coefficients?
__ S2– + __ H2O → __SO2 + __H++ __e–
__ S2– + __ H2O → __SO2 + __H++ __e–
| 1. | 2 | 2. | 4 |
| 3. | 6 | 4. | 8 |
Subtopic: Balancing of Equations |
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aMnO4–(aq) + bBr–(aq) + cH2O(l) → dMnO2(s) + eBrO3– (aq) + fOH–(aq)
The value of c, d and f in the above mentioned reaction are respectively:
The value of c, d and f in the above mentioned reaction are respectively:
| 1. | c = 1; d = 2; f = 2 | 2. | c = 2; d = 1; f = 2 |
| 3. | c = 2; d = 2; f = 1 | 4. | c = 1; d = 1; f = 1 |
Subtopic: Balancing of Equations |
67%
Level 2: 60%+
Hints
Consider the following reaction:
\({Cr}_{2}{O}_{7}^{{-}{2}}{+}{xH}^{\oplus}{+}{Fe}^{{+}{2}}\rightarrow{yFe}^{{+}{3}}{+}{2}{Cr}^{{+}{3}}{+}{zH}_{2}{O} \)
The sum of x, y, and z is:
\({Cr}_{2}{O}_{7}^{{-}{2}}{+}{xH}^{\oplus}{+}{Fe}^{{+}{2}}\rightarrow{yFe}^{{+}{3}}{+}{2}{Cr}^{{+}{3}}{+}{zH}_{2}{O} \)
The sum of x, y, and z is:
| 1. | 24 | 2. | 22 |
| 3. | 27 | 4. | 29 |
Subtopic: Balancing of Equations |
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Level 2: 60%+
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Permanganate(VII) ion, MnO4– in basic solution oxidizes iodide ion, I– to produce molecular iodine (I2) and manganese (IV) oxide (MnO2). The reaction is as follows:
aI–(aq) + bMnO4–(aq) + cH2O(l) → dI2(s) + eMnO2(s) + fOH–(aq)
The value of b, d and f are-
| 1. | b = 2; d = 3; f = 8 | 2. | b = 1; d = 3; f = 8 |
| 3. | b = 3; d = 8; f = 2 | 4. | b = 8; d = 3; f = 2 |
Subtopic: Balancing of Equations |
66%
Level 2: 60%+
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