How does adding NH3 into the cell with CuSO4 affect the voltage measurement? The other cell has ZnSO4
The addition of NH3 to a solution of copper(II) ions causes the formation of the copper ammonia complex, Cu(NH3)4^+2, and this reduces the concentration of the Cu^+2 ion in solution which reduces the voltage of that half cell and of course the voltage of the cell.
Well, adding NH3 into the cell with CuSO4 and ZnSO4 is like inviting a clown to a serious business meeting. It's going to make quite a spectacle! In this case, NH3 acts as a ligand and forms a complex with Cu2+ ions, reducing their concentration in the solution. As a result, the voltage measurement may change because there's a change in the chemical species present in the cell. So, to put it simply, adding NH3 can give our voltage measurement a rollercoaster ride, making it a bit more unpredictable and entertaining!
When NH3 is added to the cell with CuSO4 and ZnSO4 in the other cell, it results in the formation of a complex ion called [Cu(NH3)4]2+. This complex ion is more stable than Cu2+ ions, and as a result, it reduces the concentration of Cu2+ ions in the solution.
The reduction in Cu2+ ions leads to a decrease in the concentration of positive charges in the CuSO4 solution and a decrease in the overall cell voltage. This is because the cell voltage is directly proportional to the concentration of ions in the solution according to the Nernst equation.
On the other hand, in the cell with ZnSO4, the addition of NH3 has no significant effect on the voltage measurement. This is because Zn2+ ions do not form stable complexes with NH3 and their concentration remains unchanged.
In summary, adding NH3 to the cell with CuSO4 decreases the voltage measurement due to the formation of stable complex ions, while the addition of NH3 to the ZnSO4 cell has no significant effect on the voltage measurement.
To understand how the addition of NH3 (ammonia) into a cell with CuSO4 (copper sulfate) affects the voltage measurement, we need to consider the underlying electrochemical reactions and the redox potential of the involved species.
In this scenario, let's assume that the two cells are connected through a salt bridge or a porous barrier to complete the circuit.
The half-reactions occurring in the two cells are as follows:
At the anode (oxidation) of the first cell:
Cu(s) → Cu2+(aq) + 2e-
At the cathode (reduction) of the first cell:
SO42-(aq) + 4H2O(l) + 2e- → S(s) + 4OH-(aq)
At the anode (oxidation) of the second cell:
Zn(s) → Zn2+(aq) + 2e-
At the cathode (reduction) of the second cell:
2H2O(l) + 2e- → H2(g) + 2OH-(aq)
The overall reaction for the first cell is:
Cu(s) + SO42-(aq) + 4H2O(l) → Cu2+(aq) + S(s) + 4OH-(aq)
The overall reaction for the second cell is:
Zn(s) + 2H2O(l) → Zn2+(aq) + H2(g) + 2OH-(aq)
In both cells, the metal (Cu or Zn) is being oxidized (loses electrons) at the anode, and reduction reactions occur at the cathode.
When NH3 is added into the cell with CuSO4, it can form a complex with Cu2+ ions, according to the following equilibrium reaction:
Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)4^2+(aq)
By forming this complex, the equilibrium shifts towards the right (products) due to the Le Chatelier's principle, resulting in an increase in the concentration of Cu(NH3)4^2+ species.
This change in concentration affects the concentration of Cu2+ ions and reduces their presence in the solution. As a result, the oxidation reaction at the anode is slowed down because there are fewer Cu2+ ions available to be reduced.
On the other hand, in the cell with ZnSO4, NH3 does not form any significant complex with Zn2+. Therefore, adding NH3 into the ZnSO4 cell does not have a significant impact on the voltage measurement.
In summary, adding NH3 into the cell with CuSO4 affects the voltage measurement by reducing the concentration of Cu2+ ions in the solution and thus slowing down the oxidation reaction, leading to a decrease in the overall cell potential.