JEE MAIN - Chemistry (2015 (Offline) - No. 7)
Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the
cathode is: (at. mass of Cu = 63.5 amu)
63.5 g
2 g
127 g
0 g
Paskaidrojums
$$C{u^{2 + }} + 2{e^ - }\buildrel \, \over
\longrightarrow Cu$$
$$2F\,\,i.e.\,\,2 \times 96500\,C$$ deposit $$Cu=1$$ mol $$=63.5$$ $$g$$
$$2F\,\,i.e.\,\,2 \times 96500\,C$$ deposit $$Cu=1$$ mol $$=63.5$$ $$g$$