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
★☆〖Qบęຮτ ı¨ ø nˇ〗☆★
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 = 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.
____________________________________________________
<Judge It Yourself...>
Hope it helps you! ヅ
✪ Be Brainly ✪
★☆〖Qบęຮτ ı¨ ø nˇ〗☆★
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 ✪