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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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) :
MnO₄⁻(aq) + 8H⁺ (aq) + 5e⁻ → Mn²⁺ (aq) + 4H₂O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO₄⁻(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.
____________________________________________________
<Judge It Yourself...>
Hope it helps you! ヅ
✪ Be Brainly ✪