Question

Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Answer 1

Answer:

Three Electrolytic Cells A,B,C Containing Solutions Of ZnSO4, AgNO3 And CuSO4, Respectively Are Connected In Series. A Steady Current Of 1.5 Amperes Was Passed Through Them Until 1.45 G Of Silver Deposited At The Cathode Of Cell B...

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Answer 2

Final answer:

1.5 A current will flow for 14.892 minutes.

The amount of copper deposited = 0.4265 g

The amount of zinc deposited = 0.4389 g

Given that: We are given three electrolytic cells A, B, and C containing solutions of $ZnSO_{4}$, $AgNO_{3}$ and $CuSO_{4}$ respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver was deposited at the cathode of cell B.

To find: We have to find, how long the current flow and mass of copper and zinc were deposited.

Explanation:

• According to the cathode reaction in electrolytic cell B:

$Ag^{+}(aq) + e^{-}$ → $Ag(s)$

                        $108 \ g$

That is, 108 g of $Ag$ is deposited by 96500 $C$.

• Therefore, 1.45 g of $Ag$ is deposited by = $\frac{96500 \times 1.45}{108}$ = 1295.602 $C$

• In an electrolytic cell, a current strength of $I$ amperes passed for $t$ seconds, and then the quantity of electricity passed:

$Q =It\\\\t=\frac{Q}{I}$

Here:

$I$ = 1.5 $A$

$Q$ = 1295.602 $C$

Substitute corresponding values.

$t$ = $\frac{1295.602}{1.45}$ = 893.518 seconds

Time = 893.518 seconds = 14.892 minutes

• Current will flow 14.892 minutes.

• Consider the cathode reaction in electrolytic cell A:

$Zn^{2+} (aq) + 2e^{-}$ → $Zn (s)$

                          $65.38 \ g$

That is 63.54 g of zinc ($Zn$) deposited by $2 \times 96500 \ C$.

Therefore 1295.602 $C$ would deposit = $\frac{65.38}{2 \times 96500}$ $\times1295.602$ = 0.4389 g of zinc.

• Now, consider the cathode reaction in electrolytic cell C:

$Cu^{2+}(aq) + 2e^{-}$ → $Cu(s)$

                            $63.54 \ g$

That is 63.54 g of copper ($Cu$) deposited by $2 \times 96500 \ C$.

Therefore 1295.602 $C$ would deposit = $\frac{63.54}{2 \times 96500}$ $\times 1295.602$ = 0.4265 g of copper.

To know more about the concept please go through the links

https://brainly.in/question/10067832

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