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Nernst Equation

I. 2Mn2+aq ...

Question

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

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

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

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 $O_{2}$ ?

A

0.5

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B

1

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C

2

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D

4

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Solution

The correct option is D $4$
Dissolved oxygen in water is given by:
(1) $2 {M n}^{+ 2} (a q) + 4 {O H}^{-} (a q) + O_{2} (g) \to {M n O}_{2} (s) + 2 H_{2} O$
(2) $Mn O_{2} (s) + 2 I^{-} (a q) + 4 H^{+} (a q) \to {M n}^{+ 2} (a q) + I_{2} (a q) + 2 H_{2} O$
(3) $2 S_{2} O_{3}^{- 2} (a q) + I_{2}$ (aq) $⟶ S_{4} O_{6}^{- 2} (a q) + 2 I^{-}$
from eq (1) $: -$
1 mole of $O_{2}$ gives 2 moles of $M n O_{2}$
rom eq (2) :
1 mole of $M n O_{2}$ gives 1 mole of $I_{2}$
i.e. 1 mole of $O_{2}$ gives 2 mole of $I_{2}$
2 moles of $S_{2} O_{3}^{- 2}$ react with 1 mole of $I_{2}$

So, 4 moles of $S_{2} O_{3}^{- 2}$ will react with 2 moles of $I_{2}$
and 2 moles of $I_{2}$ react with 1 mole of $O_{2}$ ,
Therefore option: $D$

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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 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?

O_{2}

in water is determined by using a redox reaction as follows:

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)

How many equivalents of

O_{2}

will be required to react with

1

M n^{2 +}

?

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

2 {F e}^{3 +} (a q) + 2 I^{⊖} (a q) ⟶ I_{2} (a q) + 2 {F e}^{2 +} (a q)

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