Question

Given below are few reactions with some expressions. Mark the expression which is not correctly matched.

• A. For concentration cell,
  $Ag\left|A{g}^{+}\left(C_1\right)\right|\left|A{g}^{+}\left(C_2\right)\right|Ag$ ;
  $E_{cell}=-\frac{0.0591}{1}log\frac{C_1}{C_2}$
• B. For the cell, $2A{g}^{+}+H_2\left(1atm\right)\to 2Ag+2H^{+}\left(1M\right)$ ;
  $E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}log\frac{{\left[A{g}^{+}\right]}^2}{{\left[H^{+}\right]}^2}$
• C. For an electrochemical reaction, at equilibrium
  $aA+bB\stackrel{n{e}^{-}}{\rightleftharpoons }cC+dD$ ;
  $E_{cell}^{\circ }=\frac{0.0591}{n}log\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}$
• D. For the cell, $M_{\left(aq\right)}^{n+}+n{e}^{-}\to M_{\left(s\right)}$ ;
  $E=E^{\circ }-\frac{0.0591}{n}log\frac{1}{\left[M^{n+}\right]}$

Answer 1

Answer: (B)

For $2A{g}^{+}+H_2\left(1atm\right)\to 2Ag+2H^{+}\left(1M\right)$
$E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}log\frac{{\left[H^{+}\right]}^2}{{\left[A{g}^{+}\right]}^2}$