Question

Given the following electrolytic cell:

The current is discharged into the electrolytic cell containing the solution CuSO_4(aq) 1.0M at 25 ^oC. During the operation of the cell, copper Cu_(s) is deposited on one electrode and oxygen O2_(g) gas is released, near the second electrode.

O_2(g) + 4H₃O⁺_(aq) + 4e^-$\rightarrow$ 6H₂O_(l) E^o= 1.23V

Cu²⁺_(aq) + 2e^-$\rightarrow$ Cu_(s)    E^o= 0.34V

A. Write the direction of the flow of electrons in the cell.
B. Write the electrolysis equation that occurs in the cell.
C. Predicted if the  $\Delta G$ of reaction is negative or positive. Explain.
D. Calculate the $\Delta G$ for reaction. Show calculations.

E. The current in the cell is 1.5 amperes. The current is streamed in 40 minutes.
i. Calculate the mass of copper in grams that have sunk. Show calculations.
ii. What is the volume in liters of oxygen gas released on 25 ^oC under pressure of 1.16 atm. Show calculations.

Thanks!

Power source 2+ (aq)

Answer 1

Here, In the electrolytic cell, The non spontaneous reaction occurs.

So, The half cell reaction are-

Anode half cell reaction- 6 H₂O (l) --------->  O₂(g) + 4H₃O⁺(aq) + 4e^-   , E^o= - 1.23V

cathode half cell reaction- Cu²⁺ (aq) + 2 e^- ----------> Cu (s) ,   E^o= 0.34 V

A) Electron always flows from Anode to cathode.

Here in electrolytic cell, Anode is Positive electrode and Cathode is Negative electrode

B) The electrolysis reaction are-

6 H₂O (l) --------->  O₂(g) + 4H₃O⁺(aq) + 4e^-

Cu²⁺ (aq) + 2 e^- ----------> Cu (s)

D) $\Delta$G = $\Delta$G⁰ + RT ln Q

We know,    $\Delta$G⁰ = - n . F . E⁰_cell

So, $\Delta$G = -  n . F . E⁰_cell + RT ln Q =   -  n . F . ( -1.23 V + 0.34 V ) + RT ln Q

=>   $\Delta$G =    - n . F . ( -0.89 V  ) + RT ln Q

=>   $\Delta$G = n . F . ( 0.89 V  ) + RT ln Q

C) $\Delta$G of the reaction should be negative, for the reaction to feasible.

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