Thank you for reporting, we will resolve it shortly
Q.
If hydrogen electrode dipped in $2$ solution of $ pH=3 $ and $ pH=6 $ and salt bridge is connected the emf of resulting cell is:
Delhi UMET/DPMTDelhi UMET/DPMT 2005
Solution:
For concentration cells
$E_{\text {cell }}^{o}=\frac{0.0591}{1} \log \left[\frac{\text { anode }}{\text { cathode }}\right]$
The electrode having lower concentration acts as anode and the electrode having higher concentration acts as cathode.
The cells having same electrolytes at cathode and anode are called concentration cells.
Given: $p H=3$ for first solution and $p H=6$ for second solution. $\left[H^{+}\right]$for first solution
$=10^{-p H}=10^{-3}\left[H^{+}\right]$ for second solution
$=10^{-p H}=10^{-6}$
$\therefore E_{\text {cell }}^{o}=-\frac{0.0591}{1} \log \left[\frac{10^{-6}}{10^{-3}}\right]$
$=0.091 \log 10^{-3}=-0.0591 \times(-3 \log 10)$
$=-0.0591 \times(-3 \times 1)=-0.0591-3$
$\therefore E_{\text {cell }}^{o}=0.177\, V$