Two half-reactions of an electrochemical cell are given below: $$\mathrm{MnO_4^-(aq)+8H^+(aq)+5e^-\longrightarrow Mn^{2+}(aq)+4H_2O(l)},\quad E^{\circ}=+1.51\ \mathrm{V}$$ $$\mathrm{Sn^{2+}(aq)\longrightarrow Sn^{4+}(aq)+2e^-},\quad E^{\circ}=+0.15\ \mathrm{V}$$ Construct the redox equation from the two half-cell reactions and predict if this reaction favours formation of reactants or products shown in the equation.

CBSE Class 12 · 2010 · All India · Redox Reactions and ElectrochemistryShort Answer

Two half-reactions of an electrochemical cell are given below:

Mn O_{4}^{-} (aq) + 8 H^{+} (aq) + 5 e^{-} ⟶ M n^{2 +} (aq) + 4 H_{2} O (l), E^{\circ} = + 1.51 V

S n^{2 +} (aq) ⟶ S n^{4 +} (aq) + 2 e^{-}, E^{\circ} = + 0.15 V

Construct the redox equation from the two half-cell reactions and predict if this reaction favours formation of reactants or products shown in the equation.

Answer: 2MnO₄⁻ + 16H⁺ + 5Sn²⁺ → 2Mn²⁺ + 8H₂O + 5Sn⁴⁺; E°_cell = 1.51 − 0.15 = +1.36 V > 0 → products favoured.

Solution

Balancing electrons (×2 for Mn, ×5 for Sn):

2Mn O_{4}^{-} + 16 H^{+} + 5S n^{2 +} ⟶ 2M n^{2 +} + 8 H_{2} O + 5S n^{4 +}

E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = 1.51 - 0.15 = + 1.36 V

Since

E_{ce l l}^{\circ} > 0

(

Δ G^{\circ} < 0

), the reaction is spontaneous as written —

formation of products is favoured

CBSE Class 12 Board · 2010 · All India · Subjective

Practise more: all CBSE board PYQs on Redox Reactions and Electrochemistry →

CBSE Class 12 · 2010 · All India · Redox Reactions and ElectrochemistryShort Answer

Solution

More board questions on Redox Reactions and Electrochemistry