When copper sulphate solution is electrolysed in a copper voltameter for 30 sec, then m gram of copper was deposited. Time current graph for the electrolysis is The electrochemical equivalent of copper from above plot will be

Question

When copper sulphate solution is electrolysed in a copper voltameter for 30 sec, then m gram of copper was deposited. Time current graph for the electrolysis is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-yjcxkmwxr028.png" /> The electrochemical equivalent of copper from above plot will be (A) Z = m (B) Z=(m/2) (C) Z=(m/5) (D) Z=2m

Tardigrade Admin · Accepted Answer

m = ZIt Z=(m/I t) It = Total charge passed through copper voltameter = Area under the curve =(100 × 10 × 1 0- 3)+2((1/2) × 10 × 100 × 1 0- 3)=2 So Z=(m/2)

Q. When copper sulphate solution is electrolysed in a copper voltameter for 30 sec, then m gram of copper was deposited. Time current graph for the electrolysis is

The electrochemical equivalent of copper from above plot will be

Z = m

$Z = \frac{m}{2}$

$Z = \frac{m}{5}$

$Z = 2 m$

m = ZIt

$Z = \frac{m}{I t}$

It = Total charge passed through copper voltameter

= Area under the curve

$= (100 \times 10 \times 1 0^{- 3}) + 2 (\frac{1}{2} \times 10 \times 100 \times 1 0^{- 3}) = 2$

So $Z = \frac{m}{2}$

Q. When copper sulphate solution is electrolysed in a copper voltameter for 30 sec, then m gram of copper was deposited. Time current graph for the electrolysis is The electrochemical equivalent of copper from above plot will be

Z = m

Z=2m​

Z=5m​

Z=2m

m = ZIt Z=Itm​ It = Total charge passed through copper voltameter = Area under the curve =(100×10×10−3)+2(21​×10×100×10−3)=2 So Z=2m​

Q. When copper sulphate solution is electrolysed in a copper voltameter for 30 sec, then m gram of copper was deposited. Time current graph for the electrolysis is

The electrochemical equivalent of copper from above plot will be

$Z = \frac{m}{2}$

$Z = \frac{m}{5}$

$Z = 2 m$

m = ZIt

$Z = \frac{m}{I t}$

It = Total charge passed through copper voltameter

= Area under the curve

$= (100 \times 10 \times 1 0^{- 3}) + 2 (\frac{1}{2} \times 10 \times 100 \times 1 0^{- 3}) = 2$

So $Z = \frac{m}{2}$