Question

The molar conductances of $+ 0.19V$ and $+ 0.53V$ at infinite dilution are 126.45, 426.16 and $+ 0.30V$ respectively. The molar conductance of $Ag|Ag^{+}(C_{1})||Ag^{+}(C_{2})|Ag;E_{cell} = - \frac{0.0591}{1}\log\frac{C_{1}}{C_{2}}$ at

infinite dilution is.

• A. $2Ag^{+} + H_{2}(1atm) \rightarrow 2Ag + 2H^{+}(1M);$
• B. $E_{cell} = Eº_{Cell} - \frac{0.0591}{2}\log\frac{\lbrack Ag^{+}\rbrack^{2}}{\lbrack H^{+}\rbrack^{2}}$
• C. $aA + bB$
• D. $Eº_{cell} = \frac{0.0591}{n}\log\frac{\lbrack C\rbrack^{c}\lbrack D\rbrack^{d}}{\lbrack A\rbrack^{a}\lbrack B\rbrack^{b}}$

Answer 1

Answer: (B)

$E_{cell} = Eº_{Cell} - \frac{0.0591}{2}\log\frac{\lbrack Ag^{+}\rbrack^{2}}{\lbrack H^{+}\rbrack^{2}}$

Answer 2

$\land_{m}^{o}(CH_{3}COOH) =$

$\land^{o}(CH_{3}COONa) + \land^{o}(HCl) - \land^{o}(NaCl)$

$= 91 + 426.16 - 126.45 = 390.71ohm^{- 1}cm^{2}mol^{- 1}$.