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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 = 1.5 moles of Cu²⁺.
Thus the Nernst equation of the reaction is = E° +
log
.
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 + log[Cu²⁺]
Or, 0.029 log [Cu²⁺]= 0.246
Or, log [Cu²⁺] = 8.33
Or, [Cu²⁺] = 2.137×10⁸