CBSE Class 12 · 2004 · All India Compartment · Redox Reactions and ElectrochemistryShort Answer

Silver is electrodeposited on a metallic vessel of surface area 800 cm

^{2}

by passing a current of 0.2 A for 3 hours. Calculate the thickness of silver deposited. (Density of silver = 10.47 g cm

^{- 3}

, atomic mass of silver = 107.92 u)

Answer: ≈ 2.9 × 10⁻⁴ cm

Solution

Charge passed:

Q = I t = 0.2 \times 3 \times 3600 = 2160 C

Moles of Ag deposited (Ag

^{+}

+ e

^{-} \to

Ag):

n = \frac{2160}{96500} = 0.02238 mol

Mass:

m = 0.02238 \times 107.92 = 2.416 g

Volume:

V = \frac{2.416}{10.47} = 0.2307 c m^{3}

Thickness:

t = \frac{V}{A} = \frac{0.2307}{800}

t = 2.9 \times 1 0^{- 4} cm

CBSE Class 12 Board · 2004 · All India Compartment · Subjective