CBSE Class 12 · 2008 · Delhi · Redox Reactions and ElectrochemistryLong Answer

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Three electrolytic cells A, B and C containing solutions of zinc sulphate, silver nitrate and copper sulphate respectively are connected in series. A steady current of 1.5 ampere was passed through them until 1.45 g of silver were deposited at the cathode of cell B. How long did the current flow? What mass of copper and what mass of zinc were deposited in the concerned cells? (Atomic masses: Ag = 108, Zn = 65.4, Cu = 63.5)

Answer: t ≈ 864 s (≈ 14.4 min); Cu ≈ 0.426 g; Zn ≈ 0.439 g

Solution

Moles of Ag deposited:

n_{A g} = \frac{1.45}{108} = 0.01343 mol

Charge (Ag

^{+}

+ e

^{-} \to

Ag):

Q = 0.01343 \times 96500 = 1295.8 C

Time:

t = \frac{Q}{I} = \frac{1295.8}{1.5}

t = 864 s \approx 14.4 min

Cells in series pass the same charge. For the divalent ions (

n = 2

), moles deposited

= \frac{1295.8}{2 \times 96500} = 6.71 \times 1 0^{- 3}

mol:

m_{C u} = 6.71 \times 1 0^{- 3} \times 63.5

m_{C u} = 0.426 g

m_{Z n} = 6.71 \times 1 0^{- 3} \times 65.4

m_{Z n} = 0.439 g

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CBSE Class 12 · 2008 · Delhi · Redox Reactions and ElectrochemistryLong Answer

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