Two half cell reactions of an electrochemical cell are given below :
MnO–4(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I), E° = + 1.51 V
Sn2+ (aq) → 4 Sn4+ (aq) + 2e–, E° = + 0.15 V
Construct the redox equation from the two half cell reactions and predict if this reaction favours formation of reactants or product shown in the equation...
Answers
The reactions can be represented at anode and at cathode in the following ways :
At anode (oxidation) :
Sn2+ → = Sn4+ (aq) + 2e– ] × 5 E° = + 0.15 V
At cathode (reduction) :
MnO–4(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO–4(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O
Now E°cell = E°cathode – E°anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°cell favours formation of product.
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Answer:
Two half-reaction of an electrochemical cell are given below :
MnO
4
−
(aq)+8H
+
+5e
−
→Mn
2+
(aq)+4H
2
O(l);E
∘
=+1.51V
Sn
2+
(aq)→Sn
4+
(aq)+2e
−
,E
∘
=+0.15V
Construct the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured .
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ANSWER
At anode : [Sn
2+
(aq) → Sn
4+
(aq)+2e
−
]×5
At cathode : [MnO
4
−
(aq)+8H
+
(aq)+5e
−
→ Mn
2+
(aq)+4H
2
O(l)]×2
Cell reaction :
_____________________________________________________________________
2MnO
4
−
(aq)+5Sn
2+
(aq)+16H
+
(aq)→2Mn
2+
(aq)+5Sn
4+
(aq)+8H
2
O(l)
______________________________________________________________________
E
cell
∘
=E
cathode
∘
−E
anode
∘
=1.51V−0.15V
=1.36V