Balancing Red...

Balancing Redox Equations

Trending Questions

How many moles of $K_{2} C r_{2} O_{7}$ can be reduced by 1 mole of $S n^{+ 2}$

C_{2} H_{6} (g) + n O_{2} \to C O_{2} (g) + H_{2} O (I)

In this equation, the ratio of the coefficients of

C O_{2}

and

H_{2} O

2 : 3
1 : 1
3 : 2
1 : 3

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The number of electrons involved in the reduction of $C r_{2} O_{7}^{2 -}$ in acidic solution to $C r^{3 +}$ is

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Q. For the redox reaction

x P_{4} + y H N O_{3} \to H_{3} P O_{4} + N O_{2} + H_{2} O

x = 2, y = 10
x = 1, y = 15
x = 1, y = 5
x = 1, y = 20

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Q. Oxidation of thiosulphate

(S_{2} O_{3}^{2 -})

ion by iodine gives

$S O_{3}^{2 -}$
$S O_{4}^{2 -}$
$S_{4} O_{6}^{2 -}$
$S_{2} O_{6}^{2 -}$

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When $K M n O_{4}$ acts as an oxidizing agent and ultimately forms $M n O_{4}^{- 2}$ , $M n O_{2}$ , $M n_{2} O_{3}$ and $M n^{+ 2}$ , then the number of electrons transferred ( per Mn atom) in each case respectively is:

4, 3, 1, 5
1, 5, 3, 7
1, 3, 4, 5
3, 5, 7, 1

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Q. For decolourization of 1 mole of

K M n O_{4}

, the moles of

H_{2} O_{2}

required is

$\frac{5}{2}$
$\frac{1}{2}$
$\frac{3}{2}$
$\frac{7}{2}$

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In the reaction, $8 A l + 3 F e_{3} O_{4} \to 4 A l_{2} O_{3} + 9 F e$ , the number of electrons transferred from reductant to oxidant is:

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Q. In a balanced equation

H_{2} S O_{4} + x H I \to H_{2} S + y I_{2} + z H_{2} O

, the values of x, y, z are

x = 3, y = 5, z = 2
x = 4, y = 8, z = 5
x = 8, y = 4, z = 4
x = 5, y = 3, z = 4

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For the redox reaction
$M n O_{4}^{-} + C_{2} O_{4}^{- 2} + H^{+} ⟶ M n^{2 +} + C O_{2} + H_{2} O$
the correct coefficients of the reactants $M n O_{4}^{-}, C_{2} O_{4}^{2 -} a n d H^{+}$ for the balanced reaction are

2, 16, 5
2, 5, 16
16, 5, 2
5, 16, 2

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