CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Ion Electron Method

Using electro...

Question

Using electron-transfer concept, identify the oxidant and reductant in the following redox reactions.
(a) $Z n (s) + 2 H^{+} (a q) ⟶ Z n^{2 +} (a q) + H_{2} (g)$
(b) $2 [F e (C N)_{6}]^{3 -} (a q) + 2 O H^{-} (a q) + H_{2} O_{2} (a q) \to 2 [F e (C N)_{6}]^{4 -} (a q) + 2 H_{2} O (l)$
(c) $2 [F e (C N)_{6}]^{3 -} (a q) + 2 O H^{-} (a q) + H_{2} O_{2} (a q) \to 2 [F e (C N)_{6}]^{4 -} (a q) + O_{2} (g) + 2 H_{2} O (l)$

Open in App

Solution

(a) $Z n (s) + {2 H}^{+} (a q) \to {Z n}^{2 +} (a q) + H_{2} (g)$
oxidation :- $Z n (s) \to {Z n}^{2 +} + 2 e^{-}$
Reduction :- ${2 H}^{+} (a q) + 2 e^{-} \to H_{2} (g)$
Therefore, oxidant is $H^{+} (a q)$ & reductant is $Z n (s)$
(b) ${2 [+ 2 F e {(C N)}_{6}]}^{3 -} (a q) + {2 O H}^{-} (a q) + H_{2} O_{2} (a q) ⟶ {2 [+ 2 F e {(C N)}_{6}]}^{4 -} + 2 H_{2} O (l)$
$O^{- 1} + e^{-} \to O^{- 2}$
${F e}^{+ 3} + e^{-} \to {F e}^{+ 2}$
Here, oxidation does not take place
Therefore, $H_{2} O_{2}$ and ${[F e {(C N)}_{6}]}^{3 -}$ both are oxidants.
(c) ${2 [+ 3 F e {(C N)}_{6}]}^{3 -} (a q) + {2 - 2 O H}^{-} (a q) + H_{2} {- 1 O}_{2} (a q) \to 2 {[+ 2 F e {(C N)}_{6}]}^{4 -} (a q) + {0 O}_{2} (g) + 2 H_{2} - 2 O (l)$
Here, $+ 3 F e$ reduces to $+ 2 F e$ and $- 1 O$ oxidises to ${0 O}_{2}$ and $- 1 O$ reduces to $- 2 O$ . Therefore, ${[F e {(C N)}_{6}]}^{3 -}$ and $H_{2} O_{2}$ act as a oxidant and $H_{2} O_{2}$ act as a reductant also.

Suggest Corrections

thumbs-up

0

Similar questions

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 K₄ [Fe (CN)₆]_(aq) + H₂O_{2 (aq)} → 2 K₃ [Fe (CN)₆]_(aq) + 2KOH_(aq) is an example of

Q. Consider a spontaneous electrochemical cell between

C u

A l

. Predict what would happen if excess concentrated

N a O H

were added to the cell with copper ions and a precipitate forms.

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

Q. In the given reaction,

2 K_{4} [F e {(C N)}_{6}] (a q) + H_{2} O_{2} (a q) ⟶ 2 K_{3} [F e {(C N)}_{6}] (a q) + 2 K O H (a q)

Which of the following processes takes place?

Q. In the following reaction which element is oxidized ?

Z n (s) + 2 H^{\oplus} (a q) ⟶ {Z n}^{2 +} (a q) + H_{2} (g)

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Balancing_ion elec

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Ion Electron Method

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon