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?
According to the reaction:
Ag+(aq) + e- → Ag(s)
108 g
i.e., 108 g of Ag is deposited by 96487 C
Therefore, 1.45 g of Ag is deposited by = \(\frac{96487 \times 1.45}{108}\) C
= 1295.43 C
Given,
Current = 1.5 A
∴ Time = \(\frac{12945.43}{1.5}\) s
= 863.6 s
= 864 s
= 14.40 min
Again,
Cu2+(aq) + 2e- \(\rightarrow\) Cu(s)
63.5 g
i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu
Therefore, 1295.43 C of charge will deposit =\(\frac{63.5 \times 1295.43}{2\times 96487}\) g
= 0.426 g of Cu
Zn2+(aq) + 2e- \(\rightarrow\) Zn(s)
65.4 g
i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn
Therefore, 1295.43 C of charge will deposit = \(\frac{65.4 \times 1295.43}{2\times 96487}\) g
= 0.439 g of Zn
Standard electrode potential for \( \text{Sn}^{4+}/\text{Sn}^{2+} \) couple is +0.15 V and that for the \( \text{Cr}^{3+}/\text{Cr} \) couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be:
To calculate the cell potential (\( E^\circ_{\text{cell}} \)), we use the standard electrode potentials of the given redox couples.
Given data:
\( E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15V \)
\( E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74V \)

An electrochemical cell is a device that is used to create electrical energy through the chemical reactions which are involved in it. The electrical energy supplied to electrochemical cells is used to smooth the chemical reactions. In the electrochemical cell, the involved devices have the ability to convert the chemical energy to electrical energy or vice-versa.