Question

The photoelectric current from Na (Work function, $w_0=2.3eV$) is stopped by the output voltage of the cell $Pt\left(s\right)H_2(g,1Bar)HCl(aq.pH=1)AgClsAg\left(s\right)$. The $pH$ of aq. $HCl$ required to stop the photoelectric current form $K(w_0=2.25eV)$, all other conditions remaining the same, is _______ ${10}^{-2}$ (to the nearest integer).

Answer 1

Step 1: Given Data:

Energy of photon $=2.3–E_{cell}\left\{forNa\right\}$

Energy of photon $=2.25–E_{cell}\left\{forK\right\}$

Step 2: Calculation:

$\begin{array}{rcl}{E}_{cell}\left\{for‘Na’\right\}+0.05& =& E_{cell}\left\{for‘K’\right\}\end{array}$

Substitute the values,

$\begin{array}{rcl}0.22+0.06log\left[H^{+}\right]\left[C{l}^{–}\right]+0.05& =& 0.22+0.06log\left[H^{+}\right]\left[C{l}^{–}\right]\\ 6log(10–2)+5& =& 6log\left[H^{+}\right]\left[C{l}^{–}\right]\\ log(10–12)+log\left(105\right)& =& log{\left\{\left[H^{+}\right]\left[C{l}^{–}\right]\right\}}^6\\ {\left[H^{+}\right]\left[C{l}^{–}\right]}^6& =& {10}^{-7}\\ {\left[H^{+}\right]}^{12}& =& 10–7\\ pH& =& \frac{7}{12}=0.58\\ & =& 58\times {10}^{-2}\end{array}$

Hence, the required pH is $58\times {10}^{-2}$.