Question

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e– → Sn(s). What is the concentration of Sn2+ if Zn2+ is 2.5 × 10−3 M and the cell emf is 0.660 V? The standard reduction potentials are given below Zn+2(aq) + 2 e− → Zn(s)    E∘red == −0.76 V

Sn2+(aq) + 2 e– → Sn(s)   E∘red  −0.136 V

Answer 1

$E=0.660V$

$E^{o}_{cell}=E^{o}_{cathode}-E^{o}_{anode}=-0.136V-(-0.76)=0.624V$

$E=E^{o}-\frac{0.059V}{2}log\frac{\left [ Zn^{2+} \right ]}{\left [ Sn^{2+} \right ]}$

$\frac{-2(0.660V-0.624V)}{0.059V}=log\frac{\left [ Zn^{2+} \right ]}{\left [ Sn^{2+} \right ]}$

$log(2.5x10^{-3})-log\left [ Sn^{2+} \right ]=1.2203$

$log\left [ Sn^{2+} \right ]=-3.8224$

$\left [ Sn^{2+} \right ]=antilog(-3.8224)=1.5x10^{-4}M$