\(\therefore w=\frac{E i t}{F}=\frac{\frac{A}{n} \times i \times t}{F}\)
\(\Rightarrow \quad n=\frac{A \times i \times t}{w \times F}=\frac{118.7 \times 2 \times(5 \times 3600)}{11.1 \times 96500}=3.989\)
\(A=\) atomic weight of \(\mathrm{Sn}=118.7 \mathrm{~g} \mathrm{~mol}^{-1}\)
\(i=\text { current }=\frac{200 \mathrm{~W}}{100 \mathrm{~V}}=2 \mathrm{~A}\)
\(t=\text { time }=5 \mathrm{hrs}(5 \times 3600 \mathrm{~s})\)
\(E=\frac{A}{n}=\) equavalent weight of \(\mathrm{Sn}^{n+}\)
\(n=\) valency of \(\mathrm{Sn}\) in the salt
\(w=\) mass of \(\mathrm{Sn}\) deposited \(=11.1 \mathrm{~g}\)
Here, \(\mathrm{SnO}\) and \(\mathrm{SnO}_2\) are non-electrolytes (aqueous).
So, valency of \(\mathrm{Sn}\) will be 4 in \(\mathrm{SnCl}_4\left(\mathrm{Sn}^{4+}+4 \mathrm{Cl}^{-}\right)\).