Question

Compare the potentials between the Ag/Ag+ and Cu/Cu2+ half cells before and after adding excess NH3 to the Cu/Cu2+ half cell. Did the potential increase or decrease? Explain the result in terms of the Nernst equation.

Answer 1

Ag/Ag⁺ the potential is -0.8V

and Cu/Cu⁺² the potential is -0.34V

when ammonia is added to Ag+ solution the ions are converted to Ag(NH₃)⁺₂

the equilibrium can be written as

$Ag^{+}+2NH_{3} \Leftrightarrow Ag\left ( NH_{3} \right )_{2}^{+}$

$E_{cell}=E^{o}_{cell}+\frac{0.0591}{1}log\frac{[Ag\left ( NH_{3} \right )_{2}^{+}]}{\left [ Ag^{+} \right ]\left [ NH_{3} \right ]^{2}}$

$E_{Ag\left ( NH_{3} \right )_{2}^{+}/Ag}=E^{o}_{cell}-E^{o}_{Ag/Ag^{+}}$

$E_{Ag\left ( NH_{3} \right )_{2}^{+}/Ag}= -0.78-(-0.8)=+0.02V$

therefore electrode potential increases for Ag/Ag+ electrode.

when ammonia is added to copper halfcell then

$Cu^{+2}+4NH_{3} \Leftrightarrow Cu\left ( NH_{3} \right )_{4}^{+2}$

$E_{cell}=E^{o}_{cell}+\frac{0.0591}{2}log\frac{[Cu\left ( NH_{3} \right )_{4}^{+2}]}{\left [ Cu^{+2} \right ]\left [ NH_{3} \right ]^{4}}$

here also the potential increases.