Question

The student adds 6 m nh3 to be cuso4 solution until the cu2+ ion is essentially all converted to cu(nh3)42+ ion. the voltage of the cell, ecell, goes up to 0.92v and the cu electrode is still negative. find the residual concentration of cu2+ ion in the cell

Answer 1

Here we have to get the residual concentration of Cu²⁺ in the reaction.

The residual concentration of  [Cu²⁺] is 2.137×10⁸ m

The reaction between ammonia (NH₃) and CuSO₄ in ionic form is - Cu²⁺ + 4 NH₃ = [Cu(NH₃)₄]²⁺.

Thus 1 mole of Cu²⁺ will react with 4 moles of NH₃.

Now 6m NH₃ or 6 moles of ammonia will react with $\frac{6}{4}$ = 1.5 moles of Cu²⁺.

Thus the Nernst equation of the reaction is $E_{Cell}$ = E° + $\frac{0.059}{n}$log$\frac{Cu^{2+} }{Cu}$.

Here the number of moles is 3, as 6 moles of ammonia will react with 1.5 moles of cu²⁺. and the standard reduction potential (E°) of Cu²⁺/Cu system is (+) 0.674V and the cell potential is 0.92V

On plugging the values in the equation, we get,

0.92 = 0.674 + $\frac{0.059}{2}$log[Cu²⁺]

Or, 0.029 log [Cu²⁺]= 0.246

Or, log [Cu²⁺] = 8.33

Or, [Cu²⁺] = 2.137×10⁸

Answer 2

The residual concentration of Cu2+ ion in the cell is $2.137\times 10^8\; m$.

What is the residual concentration?

The difference between the initial concentration of a species in solution and the amount of that species in MOL in the solid phase per one liter of a saturated solution.

The reaction between copper and ammonia

$Cu^2^+ + 4 NH_3 = [Cu(NH_3)_4]^2^+.$

In the reaction, there is one mol of copper and 4 moles of ammonia

By the Nernst equation

$\rm E^\circ+ log.$

Putting the values in the equation

$0.92 = 0.674 + log[Cu^2^+]\\\\log[Cu^2^+] = 8.33$

$Cu^2^+ = 2.137\times 10^8\; m$

Thus, the residual concentration is $Cu^2^+ = 2.137\times 10^8\; m$

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