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Effect on Indicators by Acids and Bases

Write the ion...

Question

Write the ionic equation for reaction between manganese dioxide and hydrochloric acid.

A

M n O_{2} (s) + 2 C l^{-} (a q) \to M n^{2 +} (a q) + C l_{2} (g) + 2 H_{2} O (l)

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B

M n O_{2} (s) + 4 H^{+} (a q) + 2 C l^{-} (a q) \to M n^{2 +} (a q) + C l_{2} (g) + 2 H_{2} O (l)

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C

M n O_{2} (s) + 3 H^{+} (a q) + 2 C l^{-} (a q) \to M n^{2 +} (a q) + C l_{2} (g) + 2 H_{2} O (l)

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D

M n O_{2} (s) + 4 H^{+} (a q) + 2 C l^{-} (a q) \to M n^{2 +} (a q) + C l (g) + 2 H_{2} O (l)

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Solution

The correct option is A $M n O_{2} (s) + 4 H^{+} (a q) + 2 C l^{-} (a q) \to M n^{2 +} (a q) + C l_{2} (g) + 2 H_{2} O (l)$
$M n O_{2} (s) + 4 H^{+} (a q) + 2 C l^{-} (a q) \to M n^{2 +} (a q) + C l_{2} (g) + 2 H_{2} O (l)$

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Similar questions

2 M n^{2 +} (a q) + 4 O H^{-} (a q) + O_{2} (g) \to 2 M n O_{2} (s) + 2 H_{2} O (l)

M n O_{2} (s) + 2 I^{-} (a q) + 4 H^{+} (a q) \to M n^{2 +} (a q) + I_{2} (a q) + 2 H_{2} O (l)

2 S_{2} O_{3}^{2 -} (a q) + I_{2} (a q) \to S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)

Dissolved oxygen in water is determined by using a redox reaction. If the above equations describe the procedure then 8 mg dissolved oxygen will consume:

2 M n^{2 +} (a q) + 4 O H^{-} (a q) + O_{2} (g) \to 2 M n O_{2} (s) + 2 H_{2} O (l)

M n O_{2} (s) + 2 I^{-} (a q) + 4 H^{+} (a q) \to M n^{2 +} (a q) + I_{2} (a q) + 2 H_{2} O (l)

2 S_{2} O_{3}^{2 -} (a q) + I_{2} (a q) \to S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)

Dissolved oxygen in water is determined by using a redox reaction. If the above equations describe the procedure then how many moles of

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

are equivalent to each mole of

O_{2}

?

2 M n^{2 +} (a q) + 4 O H^{-} (a q) + O_{2} (g) \to 2 M n O_{2} (s) + 2 H_{2} O (l)

M n O_{2} (s) + 2 I^{-} (a q) + 4 H^{+} (a q) \to M n^{2 +} (a q) + I_{2} (a q) + 2 H_{2} O (l)

2 S_{2} O_{3}^{2 -} (a q) + I_{2} (a q) \to S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)

Dissolved oxygen in water is determined by using the given redox reactions. When

3 \times 10^{3}

O_{2}

is dissolved per litre of water, then what will be the molarity of

I^{-}

produced in the given reaction?

E_{R P}^{o}

and their respective half reactions for some changes are given below:

M n O_{4}^{-} (a q) + 8 H^{+} + 5 e \to M n^{2 +} (a q) + 4 H_{2} O (l); E^{o} = + 1.52 V

M n O_{4}^{2 -} (a q) + 4 H^{+} + 2 e \to M n O_{2} (s) + 2 H_{2} O (l); E^{o} = + 2.26 V

2 M n O_{4}^{-} (a q) + 2 e \to 2 M n O_{4}^{2 -} (a q); E^{o} = + 0.56 V

(acid med.)
(d)

M n^{3 +} + e \to M n^{2 +}; E^{o} = + 1.51 V

(acid med.)
(e)

M n^{4 +} + e \to M n^{3 +}; E^{o} = 0.95 V

The most stable oxidation state of Mn is :

Q. Consider a galvanic cell using solid

C u

F e

metals with their corresponding solutions.
What is the

E_{c e l l}^{\circ}

Standard Potential (V)	Reduction Half-Reaction
$2.87$	$F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q)$
$1.51$	$M n O_{4}^{-} (a q) + 8 H^{+} (a q) + 5 e^{-} \to M n^{2 +} (a q) + 4 H_{2} O (l)$
$1.36$	$C l_{2} (a q) + 3 e^{-} \to 2 C l^{-} (a q)$
$1.33$	$C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q) + 7 H_{2} O (l)$
$1.23$	$O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} O (l)$
$1.06$	$B r_{2} (l) + 2 e^{-} \to 2 B r^{-} (a q)$
$0.96$	$N O_{3}^{-} (a q) + 4 H^{+} (a q) + 3 e^{-} \to N O (g) + H_{2} O (l)$
$0.80$	$A g^{+} (a q) + e^{-} \to A g (s)$ $
$0.77$	$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q)$
$0.68$	$O_{2} (g) + 2 H^{+} (a q) + 2 e^{-} \to H_{2} O_{2} (a q)$
$0.59$	$M n O_{4}^{-} (a q) + 2 H_{2} O (l) + 3 e^{-} \to M n O_{2} (s) + 4 O H^{-} (a q)$
$0.54$	$I_{2} (s) + 2 e^{-} \to 2 I^{-} (a q)$
$0.40$	$O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)$
$0.34$	$C u^{2 +} (a q) + 2 e^{-} \to C u (s)$
$0$	$2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)$
$- 0.28$	$N i^{2 +} (a q) + 2 e^{-} \to N i (s)$
$- 0.44$	$F e^{2 +} (a q) + 2 e^{-} \to F e (s)$
$- 0.76$	$Z n^{2 +} (a q) + 2 e^{-} \to Z n (s)$
$- 0.83$	$2 H_{2} O (l) + 2 e^{-} \to H_{2} (g) + 2 O H^{-} (a q)$
$1.66$	$A l^{3 +} (a q) + 3 e^{-} \to A l (s)$
$- 2.71$	$N a^{+} (a q) + e^{-} \to N a (s)$
$- 3.05$	$L i^{+} (a q) + e^{-} \to L i (s)$

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