Balancing Redox Equations
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In the balanced chemical reaction,
IO−3+a I−+b H+⟶c H2O+d I2
a, b, c and d respectively correspond to
5, 6, 3, 3
5, 3, 6, 3
3, 5, 3, 6
5, 6, 5, 5
In this equation, the ratio of the coefficients of CO2 and H2O is
- 1 : 1
- 2 : 3
- 3 : 2
- 1 : 3
- SO2−3
- SO2−4
- S4O2−6
- S2O2−6
MnO−4(aq)+I−(aq)→MnO2(s)+I2(s)
- 12
- 32
- 52
- 72
(ii) MnO−4(aq)+H2O2(aq)→
(iii) CaO(s)+H2O(g)→
(iv) AlCl3(g)+H2O(l)→
(v) Ca3N2(s)+H2O(l)→
(a) Lead(II) chloride reacts with Cl2 to give PbCl4.
(b) Lead(IV) chloride is highly unstable towards heat.
(c) Lead is known not to form an iodide, PbI4.
whenSO2 is passed in a solution of potassium iodate, the oxidation state of iodine change's from :
+5 to 0
-5 to 0
-7 to -1
+5 to -1
Oxidation number of carbon in C3O2 , Mg2C3, are respectively
-4/3, +4/3
+4/3, -4/3
-2/3, +2/3
-2/3, +4/3
In the redox reaction
xMnO +yPbO2+zHNO3⟶HMnO4 +Pb(NO3)2 +H2O
X = 2, Y =5, Z= 10
x =2, y = 7, z=8
x= 2, y =5 , z=8
x=2, y =5 , z=5
Consider the following reaction
xMnO−4+yC2O2−4+zH+⟶Mn+2+2yCO2+z2H2O
the values of x, y and z in the reaction are, respectively:
2, 5 and 16
5, 2 and 8
5, 2 and 16
2, 5 and 8
- 12 of molecular weight
- Molecular weight
- 14 of molecular weight
- None
- +2, 0
- +4, 0
- +2, −2
- None of the above
(a) a solution of Na2CO3 is alkaline ?
(b) alkali metals are prepared by electrolysis of their fused chlorides ?
(c) sodium is found to be more useful than potassium ?
MnO−4+C2O2−4+H+→Mn2+CO2+H2O
The correct coefficients of the reactants for the balanced equation are:
- MnO−4 - 2, C2O2−4 - 16, H+ - 5
- MnO−4 - 2, C2O2−4 - 5, H+ - 16
- MnO−4 - 16, C2O2−4 - 5, H+ - 2
- MnO−4 - 5, C2O2−4 - 16, H+ - 2
- Na2SO4
- Al(NO3)3
- NaCl
- Al2(SO4)3
H2SO4+Na2CO3→Na2SO4+H2O+CO2
- 2.46 L
- 24.60 L
- 1.27 L
- 12.30 L
xMnO−4+yC2O2−4+zH+→xMn2++2yCO2+z2H2O
The values of x, y and z in the reaction are, respectively:
- 5, 2 and 16
- 2, 5 and 8
- 2, 5 and 16
- 5, 2 and 8
XH2SO4+YSO2+ZNa2Cro4 ⟶Cr2(SO4)3) +ANa2SO4+BH2OX, Y, Z, AandB are?
2, 3, 2, 2, 2
2, 2, 2, 2, 2
4, 3, 2, 1, 0
0, 1, 2, 3, 4
- MnO−4C2O2−4H+2516
- MnO−4C2O2−4H+1652
- MnO−4C2O2−4H+5162
- MnO−4C2O2−4H+2165
H2O2+2KMnO4+3H2SO450% yield−−−−−→K2SO4+2MnSO4+3O2+4H2O
150mL of H2O2 sample was divided into two parts. The first part was treated with KI and the formed KOH required 200 mL of M/2 H2SO4 for neurtralisation. The other part was treated with KMnO4, which yielded 6.74 litre of O2 at 1 atm and 273 K. Using the percentage yield indicated, find the volume strength of the H2O2 sample used.
- 5.04
- 10.08
- 3.36
- 33.6
- One
- Two
- Three
- Four
Na2O+H2O→
- 125
- 110
- 120
- 150