CBSE Class 12 · 2010 · Redox Reactions and ElectrochemistryLong Answer

a) How many moles of mercury will be produced by electrolysing a 1.0 M Hg(NO

_{3}

)

_{2}

solution with a current of 2.00 A for 3 hours? [Hg(NO

_{3}

)

_{2}

= 200.6 g mol

^{- 1}

]
b) A voltaic cell is set up at 25°C with the half-cells Al

^{3 +}

(0.001 M) | Al and Ni

^{2 +}

(0.50 M) | Ni. Write the equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
(Given:

E_{N i^{2 +} /Ni}^{\circ} = - 0.25

E_{A l^{3 +} /Al}^{\circ} = - 1.66

Answer: (a) ≈ 0.112 mol Hg (b) 2Al + 3Ni²⁺ → 2Al³⁺ + 3Ni; E_cell ≈ 1.46 V

Solution

(a) Charge:

Q = 2.00 \times 3 \times 3600 = 21600 C

^{2 +}

+ 2e

^{-} \to

Hg, so

n_{H g} = \frac{21600}{2 \times 96500}

n_{H g} = 0.112 mol

(b) Al (lower E°) is the anode:

2Al (s) + 3N i^{2 +} (aq) ⟶ 2A l^{3 +} (aq) + 3Ni (s)

E_{ce l l}^{\circ} = - 0.25 - (- 1.66) = 1.41 V, n = 6

E_{ce l l} = 1.41 - \frac{0.059}{6} lo g \frac{( 1 0 ^{- 3} ) ^{2}}{( 0.50 ) ^{3}} = 1.41 + 0.050

E_{ce l l} = 1.46 V

CBSE Class 12 Board · 2010 · Subjective