Question
One half cell in a voltaic cell is constructed by dipping silver rod in ${AgNO}_{3}$ solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of ${ZnSO}_{4}$.
A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of ${Ag}^{+}$ions used in terms of $logx\left( x = \left\lbrack {Ag}^{+} \right\rbrack \right)$?
$E_{{Zn}^{2 +}/Zn}^{\Theta} = - 0.76\text{ }V,E_{{Ag}^{+}/Ag}^{\Theta} = + 0.80\text{ }V$, $\frac{2.303RT}{F} = 0.059\text{ }V$
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Solution
Sol. [Key Concept: Nernst: $E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n}log\ Q$ with $E^{\circ} = 0.80 - ( - 0.76)$ and $Q = \frac{1}{\lbrack{Ag}^{+}\rbrack^{2}}$; solve for $log\lbrack{Ag}^{+}\rbrack$.]
Cell Reaction :
$${Zn(s) + 2{Ag}^{\oplus}(aq) \longrightarrow 2Ag(s) + {Zn}^{+ 2}(aq) }{E_{\text{cell}\text{~}} = E_{\text{cell}\text{~}}^{o} - \frac{0.059}{n}log(Q) }{1.6 = 1.56 - \frac{0.059}{2}log\frac{1}{\left( {Ag}^{+} \right)^{2}} }{0.04 = + 0.059log\left\lbrack {Ag}^{+} \right\rbrack }{log\left\lbrack {Ag}^{+} \right\rbrack = \frac{4}{5.9} }$$
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