If the cathode is a Hg electrode, the maximum weight (in gram) of amalgam formed from this solution is

Question

If the cathode is a Hg electrode, the maximum weight (in gram) of amalgam formed from this solution is (A) 200 (B) 225 (C) 400 (D) 446

Tardigrade Admin · Accepted Answer

At cathode: Na+ + e- xrightarrow Hg Na(Hg) Amalgam Two moles of Na formed during electrolysis would produce two moles of Na(Hg) amalgam. ⇒ mass of amalgam = 2 × (23+200)=446 g

Q. If the cathode is a $H g$ electrode, the maximum weight (in gram) of amalgam formed from this solution is

200

225

400

446

At cathode : $N a^{+} + e^{-} H g N a (H g)$
$_{A ma l g am}$
Two moles of Na formed during electrolysis would produce two
moles of Na(Hg) amalgam.
$\Rightarrow$ mass of amalgam $= 2 \times (23 + 200) = 446 g$

Q. If the cathode is a Hg electrode, the maximum weight (in gram) of amalgam formed from this solution is

200

225

400

446

At cathode : Na++e−Hg​Na(Hg) Amalgam​ Two moles of Na formed during electrolysis would produce two moles of Na(Hg) amalgam. ⇒ mass of amalgam =2×(23+200)=446g

Q. If the cathode is a $H g$ electrode, the maximum weight (in gram) of amalgam formed from this solution is

At cathode : $N a^{+} + e^{-} H g N a (H g)$
$_{A ma l g am}$
Two moles of Na formed during electrolysis would produce two
moles of Na(Hg) amalgam.
$\Rightarrow$ mass of amalgam $= 2 \times (23 + 200) = 446 g$