A current of $1 \times 10^{- 5} A$ is passed through a solution of single atom thick layer of $A g N O_{3}$ for 16 min 5s, using plantinum electrodes. A uniform single atom thick layer of Ag is deposited covering 90% of cathode surface. The total surface area ( $i n c m^{2}$ ) of cathode if each silver atom covers an area of $5.6 \times 10^{- 16} c m^{2}$ is________.

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Solution

Number of Faradays = $\frac{10^{- 5} \times 965}{96500} = 1 \times 10^{- 7} F$
Number of atoms deposited = $6.023 \times 10^{23} \times 1 \times 10^{- 7}$
$= 6.023 \times 10^{16}$
Let total surface area be S.
$\frac{90}{100} \times S = 6.023 \times 10^{16} \times 5.6 \times 10^{- 16}$
$\Rightarrow S = 37.48 c m^{2}$

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Number of Faradays = 10−5×96596500=1×10−7FNumber of atoms deposited = 6.023×1023×1×10−7=6.023×1016Let total surface area be S.90100×S=6.023×1016×5.6×10−16⇒S=37.48cm2